Assume there are only finitely many such primes, and let them be $p_{1}, \cdots, p_{k}$. We consider $\left(2 p_{1} \cdots p_{k}\right)^{2}+1=P$. By the assumption and $P \equiv 1(\bmod 4)$, we know that $P$ is not a prime. Let $p$ be a prime factor of $P$, $p$ is of course odd, so $-1$ is a quadratic residue modulo $p$, i.e., $\left(\frac{-1}{p}\right)=1$. By Theorem 1, we know $p \equiv 1(\bmod 4)$, but it is clear that $p \neq p_{j}(1 \leqslant j \leqslant k)$, which contradicts the assumption. Proof completed.