Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 5 - Objective Problem 3

I Really do not understand which primes satisfy the condition.

If $p=2$ , then $3.2^{q-1}+1 | 11^2+7^2=170$ . Therefore $q \leq 6$ and for non of such $q$ the condition is satisfied.

If $p \neq 2$ , then $p$ is odd and $11^p+7^p=(11+7)(11^{p-1}-11^{p-2}.7+11^{p-3}.7- \dots +7^{p-1} )$ . So, $4 \nmid 11^p+7^p$ . But, if $q$ is odd then $4 | 3.p^{q-1}+1$ , because (knowing Euler theorem)

$3.p^{q-1}\equiv -1.(p^{2})^k \equiv -1 \ mod 4$

So, for no prime pair $(p,q)$ the condition is satisfied.