SAMC

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

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

Case 2. If $p\ge 3$, then we can write, and using Fermat Little Theorem, $$q^p=2p^q-7=2\left(p^q-p\right)+(2p-7)\Rightarrow q\mid 2p-7$$ and $$2p^q=q^p+7=\left(q^p-q\right)+(q+7)\Rightarrow p\mid q+7.$$ Set $q+7=kp$. If $2p-7\le 0\Rightarrow p=3$, hence $q\mid -1$, not possible. If $2p-7>0\Rightarrow 2p-7\ge q$, hence $2p\ge q+7=kp$, and we get $k\le 2$. 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 $p,q\ge 3$. Hence $7=2p^q-q^p=p^q\ge 27$, 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 $(p,q)=(2,3),(5,3)$.