Could this be due to the fact that I left the PGOOD pin opened?
No. That is an optional output. Leaving it unconnected won't cause a problem.
When I then noticed my error, I connected the GND pin properly and it instantly shorted with a big spark at the GND pin.
Until you have enough experience to know when to break it, a good rule of thumb is: Never make changes to a circuit with power still applied.
Even if you are changing to the correct connections, doing so with power applied can cause partial (sometimes hidden) or fatal damage to components. So please, despite the temptation to fix the connection problem quickly when you notice it, power off (and wait for any capacitors to discharge) first! :-)
The spark suggests a large potential difference between the ground at the output (which you connected at the time you saw the spark?) and the ground of your regulator. Although I can't say that this definitely caused damage, it won't have done any good to anything to equalise the potentials so quickly.
Regarding the PGOOD pin:
I'm not sure how to apply this as I do not have a DC voltage at 7V in my circuit.
That topic is answered in the datasheet, section 7.3.9:
Voltage seen by the PGOOD pin should never exceed 7 V. A resistor divider pair can be used to divide the voltage down from a higher potential.
Therefore if you were actually going to use that PGOOD signal, you could use a resistor divider from the 9V output (once your regulator is supplying a stable output!) to generate whatever pull-up voltage you require. However if you are leaving that pin unconnected, it sounds like perhaps you aren't using that signal, in which case you simply don't use it.
I first forgot to connect the GND and obtained the desired voltage, but oscillating, as my arduino was turning on and off rapidly.
Personally, if I'm designing a power supply, I don't use something valuable connected to the regulator's output while I'm testing, in case of unexpected problems which might cause damaging output voltages. Instead of using your Arduino, you may want to consider making a "disposable" load e.g. car light bulbs, or high-wattage resistors. There are also robust electronic loads available.
Finally, I wanted to point out that this IC requires special care regarding the PCB layout (see datasheet section 10) and component choice (inductor, diode and capacitors, see datasheet section 8). If you are trying to make this using some kind of breadboard arrangement, you may not be successful - even if it seems to work, it may not be stable, and/or may emit lots of EMI :-(
Update: As discussed in later comments, TI offer evaluation module LMR16030PEVM
for this IC, which provides a "known working" PCB, and was a successful solution in this case.