There is nothing inherently 'magical' about performing a power of 2 DFT, other than the fact that performing a power of 2 DFT allows one to perform the DFT in $O(Nlog(N))$ instead of $O(N^2)$. So the power of 2 DFT, (The algorithm that does this is known as the FFT), allows you to simply speed up your DFT computation by a huge factor.
I apply the fft again and the bin number changes (which is normal and
it is where I expect it to be), the amplitude is the same but the
phase angle is different) first is this normal?
If you do a larger DFT than your data vector, you are essentially going to be interpolating in the frequency domain. Thus, your new peak might not be the old equivalent peak that you first detected, before you took a larger DFT. And since it is not the same, you are essentially choosing a different complex exponential (sine plus cosine) basis this time around, meaning you would likely have a different phase value, yes.
PS: neither of the set ups (mentioned above) give data of length of
power of 2, say the first one gives 1620 data points and the second
one gives 1745 data points, so should be taking the next power of 2
for both from the beginning?
Yes, if you want to take a power of 2 FFT, then you would simply chose the next power of 2 length FFT that is larger than your data record length.
i dont necessarily want or not want to take the power of 2 FFT (time
performance is not my issue at all), more like, do I need to?
You should never take an FFT of length less than your record length, unless you want to discard data. The question of "How big does my FFT need to be", assuming the FFT length is larger than your data record length, then quickly becomes application dependent. Usually you can get away with an FFT length the same as your record length. However, sometimes you want to pick a peak from a 'smoother' FFT. In this case, you can take a larger FFT length, (2 times more, 3 times more, 10 times more, etc), and you would have interpolated your peak in the frequency domain. There is no magic number, however. Remember that the granularity of your FFT result is always $\frac{f_s}{N}$.