A Pair of Primes

$q\mid \left(2^p-1\right)\left(2^{2p}+2^{p}+1\right)=2^{3p}-1=8^{p}-1$

And by Fermat's little theorem $q\mid 2^{q-1}-1$ , since $\gcd(q,2)=1$ since $q>7$ . Same for $q\mid 8^{q-1}-1$ .

Then $q\mid 2^{\gcd(3p,q-1)}-1$ and $q\mid 8^{\gcd(p,q-1)}-1$ .

Assume for contradiction $\gcd(p,q-1)=1$ . Then $q\mid 8^1-1=7$ , impossible since $q>7$ . Therefore $\gcd(p,q-1)=p$ .

Assume for contradiction $\gcd(3p,q-1)=p$ . Then $q\mid 2^p-1$ . But $q\mid 2^{2p}+2^p+1=\left(2^p-1\right)^2+3\cdot \left(2^p-1\right)+3\iff q\mid 3$ , impossible since $q>7$ . Therefore $\gcd(3p,q-1)=3p$ , so $q-1=3pk$ for some $k\in\Bbb Z_{\ge 1}$ . $q-1$ is even and $3p$ is odd, so $q-1=6pm$ for some $m\in\Bbb Z_{\ge 1}$ .