If this is not true, then let's assume all such primes are $p_1$, $p_2$, $\cdots$, $p_k$. Clearly, $k \ge 1$ because $5$ is such a prime. Then consider the following number $$N=(p_1p_2\cdots p_k)^2+1$$ By assumption, $N$ cannot be a prime. First, it cannot be in the form of $2^m$ because $$N = (p_1p_2\cdots p_k)^2+1 \equiv 1^2 + 1\equiv 2\pmod{4}$$ and clearly, $N$ is greater than $5$ because one of $p_i$ is 5. Thus, $N$ must have at least one odd prime. Meanwhile, because $N$ is in the form of $(a^2 +1)$, all its odd prime must be in the form of $(4k+1)$. (See %%HREF%%3870%%.) However, $N$ cannot be divisible by any $p_i$. This leads to a contradiction.