Task: Let p and q be prime numbers with $p \equiv 3 \ \pmod 4$ and $q=2p+1$. Prove that $2^{p} - 1$ is prime only when $p=3$.
This question has been asked on this forum for a long time, but I have some questions about the solution that was proposed. That's why I asked a new question, because I'm not sure if someone answered me under the comments of a post six years ago. (that post is right here)
Do I understand correctly that in the solution proposed by reuns, it was proved that $q$ divides $2^p - 1$ when $p \equiv 3 \ \pmod 4$, $q \equiv 7 \pmod 8$? And in this case, the number $2^p - 1$ will be composite, not prime. But then I don't understand at all, because when $p = 3: p \equiv 3 \pmod 4$, $q \equiv 7 \pmod 8$, but the Mersenne number $2^p - 1 = 2^3 - 1 = 7$ will be prime, not composite.
And is it necessary to prove that $q \neq 2^p - 1$, when $p \neq 3$? If so, how to prove it?
If it will be easier for you not to answer my questions, but to write a solution to this task in more detail than reuns, then I will be very grateful to you.