It may be worth slogging through the exact math to illustrate what's happening here. Assuming a source resistor \$R_S\$ and a load resistor \$R_L\$ the current through both resistors will simply be
$$I = \frac{V}{R_S+R_L}$$
The power over the load becomes
$$P = I^2 \cdot R_L = V^2\frac{R_L}{(R_S+R_L)^2}$$
If \$R_L\$ is very small the power becomes zero because the numerator becomes 0. If \$R_L\$ is very large the power also becomes zero since the \$R_L^2\$ term in the denominator dominates.
Let's take a look at a graph for a source of 1 Volt/1 Ohm:
Even though the source can deliver a maximum of 1 Watt, the max we can get into the load is 0.25 Watts.
To maximize the power, we need maximized this function, i.e. solve for
$$\frac{\partial P}{\partial R_L} = 0$$
Unfortunately that's bit of a slog
$$0 = \frac{\partial P}{\partial R_L} = V^2\frac{(R_S+R_L)^2-2R_L(R_S+R_L)}{(R_S+R_L)^4} $$
$$0 = (R_S+R_L)^2-2R_L(R_S+R_L) = R_S^2+2R_SR_L+R_L^2-2R_LR_S-2R_L^2 = R_S^2-R_L^2$$
If we only the allow positive resistances, we clear see that the only solution is
$$R_L = R_S$$
If we would allow for complex impedances, we would get \$Z_L = Z_S^*\$, i.e. load and source impedance have the same magnitude but opposite phases. We leave this derivation for another day :-)