Question about the solutions to quadratic congruence $x^2\equiv -1(\mbox{mod}\;p)$

Question

As is known to all, when $p\equiv 1(\mbox{mod}\; 4)$, there are 2 solutions to the congruence in the set $\{1,2,3,...p-1\}$: $$x^2\equiv-1(\mbox{mod}\;p)$$ which to be exact are $\pm\frac{p-1}{2}!(\mbox{mod}\;p)$ My question is, are there infinitely many $p$ such that at least one of the $2$ solutions is a prime.