If $p,q$ are both odd or both even, then the expression is even and greater than $2$. Thus WLOG $q=2$ and $p$ is odd. Then $p^2+8$ must be prime. If $p=6k\pm 1$, then $p^2+8=36k^2\pm 12k+9=3(12k^2\pm 4k+3)$. Since the expression in the parentheses can't be equal to $1$, which is easily shown, we get that $p^2+8$ is composite. Therefore $p$ must be odd and can't be representable as $6k\pm 1$. The only such prime is $p=3$, and $3^2+8$ is prime. Thus the only solutions are $(p,q)\in\{(2,3),(3,2)\}$.