Saudi Arabian Mathematical Competitions

It is clear that $q$ must be odd, hence $q \geq 3$.

Case 1. If $p = 2$, then we get $2^{q+1} = q^{2} + 7$. For any $n \geq 4$ we have $2^{n+1} > n^{2} + 7$ (by induction). Hence $q = 3$, and we get solution $(2, 3)$.

Case 2. If $p \geq 3$, then we can write, and using Fermat Little Theorem,
$$q^{p} = 2 p^{q} - 7 = 2\left(p^{q} - p\right) + (2p - 7) \Rightarrow q \mid 2p - 7$$
and
$$2 p^{q} = q^{p} + 7 = \left(q^{p} - q\right) + (q + 7) \Rightarrow p \mid q + 7.$$
Set $q + 7 = k p$.
If $2p - 7 \leq 0 \Rightarrow p = 3$, hence $q \mid -1$, not possible.
If $2p - 7 > 0 \Rightarrow 2p - 7 \geq q$, hence $2p \geq q + 7 = k p$, and we get $k \leq 2$. Therefore, we have only two possibilities for $k$: $k = 1$, $k = 2$.
If $k = 1 \Rightarrow q + 7 = p \Rightarrow q \mid 2q + 7 \Rightarrow q = 7$ and $p = 14$, not a prime, contradiction.
If $k = 2 \Rightarrow q + 7 = 2p$. If $q > p$, then we have $p, q \geq 3$. Hence $7 = 2 p^{q} - q^{p} = p^{q} \geq 27$, since $p^{q} > q^{p}$, not possible. Then $p > q$, and we get $q + 7 = 2p > 2q \Rightarrow q < 7 \Rightarrow q = 3$ or $q = 5$. In first case we obtain $p = 5$ and in the second case contradiction. Finally, the solutions are $(p, q) = (2, 3), (5, 3)$.