Why does collector resistance \$R_C\$ affect base current? The answer is it doesn't, at least not much. You've just got something wrong somewhere. There's always a slight dependence of anything on everything else, so nothing in the universe is truly independent, and you can expect some fluctuation, but base current is largely independent of collector current.
It sounds like you are lost. The answer lies in the maths, and you have to do the maths to understand. It's OK to guess resistor values, and learn from your mistakes, but it's better to start on a solid foundation of understanding. So, what follows is adapted from a primer & execise sheet I wrote for my students on day one of transistor stuff.
The transistor can be switched fully off (cut-off), fully on (saturated) or somewhere in between. Considering only the fully-on or fully-off states, for the moment, we can get some valuable information to use later. Pretend that the transistor is a switch, and ignore anything happening at the base for now:
simulate this circuit – Schematic created using CircuitLab
Middle and right I emulate the the transistor conditions cut-off
(middle, where it behaves like an open switch) and saturated (right, where it behaves like a closed switch). Collector current is measured by ammeters. Obviously with the switch open (cut-off) no current can pass:
$$ I_{C(CUT)} = 0 $$
However, the right-hand "saturated" model demonstrates that current can never exceed some maximum, predicted by Ohm's law:
$$ I_{C(SAT)} = \frac{5V}{R_C} = 50mA $$
If you wanted some other saturation current, you use that same equation to find an appropriate resistance. Next we consider base current, and how we'll calculate and control it:
simulate this circuit
When this transistor is saturated or operating in its linear region, the potential difference between its base and emitter will be about 0.7V. It's actually a little more here, and it can vary between 0.5V and 0.9V depending on various things like transistor model and base current, but people generally say \$V_{BE}\approx 0.7V\$. We'll do the same here.
Kirchhoff's Voltage Law tells us that the sum of \$V_{BE}\$ and the voltage \$V_{RB}\$ across the base resistor must be the same as the difference between the potentials at each end:
$$
\begin{aligned}
V_{RB} + V_{BE} &= V_{IN} - 0V \\ \\
V_{RB} + 0.7V &= +5V - 0V \\ \\
V_{RB} &= +5V - 0.7V \\ \\
&= 4.3V
\end{aligned}
$$
Using Ohm's law again, we can predict the current through \$R_B\$, which ends up entering the base, and which we call "base current" \$I_B\$:
$$
\begin{aligned}
I_B &= \frac{V_{RB}}{R_B} \\ \\
&= \frac{4.3V}{1k\Omega} \\ \\
&= 4.3mA
\end{aligned}
$$
The simulation shows 4.3mA, the difference being due to the estimation \$V_{BE}=0.7V\$. Otherwise everything checks out.
Now let's put everything together and relate these two currents \$I_C\$ and \$I_B\$:
simulate this circuit
In the books it says:
$$ I_C = \beta \times I_B $$
My simulated transistor has a current gain \$\beta\$ (sometimes called \$h_{FE}\$) \$\beta = 140\$. From that you might therefore assume that:
$$
\begin{aligned}
I_C &= 140 I_B \\ \\
&= 140 \times 4.3mA \\ \\
&= 0.6A
\end{aligned}
$$
That's impossible, as we already established that collector current can't exceed \$I_{C(SAT)}=50mA\$. Therefore, this transistor is completely and utterly saturated, and passing the maximum collector current possible. If you look closely at the simulated circuit above, you can see that collector-to-emitter voltage \$V_{CE}\$ is slightly above 0V, which tells you that a bipolar junction transistor is not a perfect switch. A perfect 0Ω switch would have zero volts across it. We generally assume that a saturated transistor will have \$V_{CE}=0.1V\$, or for some devices \$V_{CE}=0.2V\$. Because of this, our estimated \$I_{C(SAT)}\$ was a little optimistic, but that's of little consequence here, we were close enough.
Let's try to take the transistor out of saturation. There are several ways to do this:
Reduce \$R_C\$. This will increase \$I_{C(SAT)}\$, and if we increase \$I_{C(SAT)}\$ to well over 0.6A, that would work.
Reduce \$V_{IN}\$. This would reduce base current \$I_B\$.
Increase \$R_B\$. This would also reduce \$I_B\$.
Since you are using a fixed \$V_{IN}=+5V\$, I'll try to find a value for \$R_B\$ that will place the transistor half-way between cut-off and saturation, which occurs at:
$$ I_C = \frac{I_{C(SAT)}}{2} = 25mA $$
First find the corresponding base current necessary in this state:
$$
\begin{aligned}
I_C &= \beta I_B \\ \\
I_B &= \frac{I_C}{\beta} \\ \\
&= \frac{25mA}{140} \\ \\
&= 180\mu A
\end{aligned}
$$
There's no point using figures with more than two significant digits, since everything is somewhat approximate. We use the KVL equation from before, together with Ohm's law, to obtain a suitable resistance \$R_B\$:
$$
\begin{aligned}
I_B &= \frac{V_{IN}-V_{BE}}{R_B} \\ \\
R_B &= \frac{5V - 0.7V}{I_B} \\ \\
&= \frac{4.3V}{180\mu A} \\ \\
&= 24k\Omega \\ \\
\end{aligned}
$$
Replace the 1kΩ resistor with 24kΩ:
simulate this circuit
That's perfect. We have 25mA of collector current, and \$V_{CE}\$ is now half of the supply.
The next question is how to know the \$\beta\$ of your own transistor in real life, it's crucial for all those calculations. Glad you asked. You don't know, until you measure it, so measure it. Start with setting base current to something small, like 100μA:
$$
\begin{aligned}
R_B &= \frac{5V - 0.7V}{I_B} \\ \\
&= \frac{4.3V}{100\mu A} \\ \\
&= 43k\Omega \\ \\
\end{aligned}
$$
That's a hard resistance to find, so let's use 47kΩ instead:
$$
\begin{aligned}
I_B &= \frac{4.3V}{47k\Omega} \\ \\
&= 91\mu A \\ \\
\end{aligned}
$$
Now choose a collector resistor \$R_C\$. You could guess and try different values, but the intelligent approach is to assume high current gain \$\beta=300\$, and calculate the resistance that would develop 2V or so with the corresponding collector current passing through it:
$$
\begin{aligned}
I_C &= \beta I_B \\ \\
&= 300 \times 91\mu A \\ \\
&= 27mA \\ \\
R_C &= \frac{2V}{27mA} \\ \\
&= 74\Omega \\ \\
\end{aligned}
$$
74Ω is another tough value to find, so I'll use 68Ω instead. It's not important to be exact here. Build the circuit, and measure the voltage across \$R_C\$:
simulate this circuit
To find \$\beta\$ we need to know both base current \$I_B\$ and collector current \$I_C\$. Base current has been set already, \$I_B=91\mu A\$, but collector current can be calculated from our measured voltage across \$R_C\$, using Ohm's law:
$$
\begin{aligned}
I_C &= \frac{V_{RC}}{R_C} \\ \\
&= \frac{1.32V}{68\Omega} \\ \\
&= 19mA \\ \\
\end{aligned}
$$
Now we find the ratio \$\beta\$ of the two currents:
$$
\begin{aligned}
I_C &= \beta I_B \\ \\
\beta &= \frac{I_C}{I_B} \\ \\
&= \frac{19mA}{91\mu A} \\ \\
&= 209
\end{aligned}
$$
