Divisors of Mersenne numbers

Order. $2^p\equiv1\pmod q$, so ord$_q(2)\mid p$, hence ord$_q(2)\in\{1,p\}$ (since $p$ prime). If ord $=1$, $q\mid1$, impossible. So ord $=p$.

Fermat. ord$_q(2)\mid q-1$, so $p\mid q-1$, i.e. $q\equiv1\pmod p$.

Mod 8. $q$ odd prime, $q\equiv1,3,5,7\pmod8$. Since $q\mid2^p-1$ and $2^p\equiv1\pmod q$ with $p$ odd: $q$ must be $\pm1\pmod8$ (by quadratic reciprocity / Euler's criterion, $2^{(q-1)/2}\equiv(q/2)\pmod q$).

Conclusion. $q\equiv1\pmod p$ and $q\equiv\pm1\pmod8$.