5. Assume there are only finitely many such primes, let them be $p_{1}, p_{2}, \cdots, p_{k}$.

We consider $\left(2 p_{1} \cdots p_{k}\right)^{2}+1=p$. By assumption and $p \equiv 1(\bmod 4)$, so $p$ is not a prime, let $p_{0}$ be a prime factor of $p$, $p_{0}$ is of course odd, so -1 is a quadratic residue modulo $p_{0}$, i.e., $\left(\frac{-1}{p_{0}}\right)=1$, thus $p_{0} \equiv 1(\bmod 4)$, but $p_{0}$ is clearly not $p_{1}, p_{2}, \cdots, p_{k}$, which contradicts the assumption.
Therefore, there are infinitely many primes of the form $4 k+1$.