If $q$ is not even, then $p+1$ must be a perfect square, so $p=x^2-1=(x-1)(x+1)$ for some positive integer $x$, and so if $x\geq 3$, then $p$ can be factored, contradiction, so $x=2\implies p=3$. If $q=2$, then the expression is obviously a perfect square, so our answer is $\boxed{(p,q)=(3,r), (r,2)}$, where $r$ is any prime.