Solution by Matin Yousefi, Mohammad Sharify: The part that $S$ is infinite follows from this. And to prove $P-S$ is finite take a prime $p \equiv 3 \mod 8 , p\equiv 1 \mod 3$ then both $2, 3$ are non-quadratic residues modulo $p$ and since one of $n, n^2+1$ is odd and the other is even this cannot happen.

Another way to generate an infinite sequence of primes $(p_i)_{i\geq 1}$ in $S$ is by taking $n=\prod_{k=1}^m \phi(p_k)$. Then $2^{n^2+1}-3^n$ isn't divisible by any of $p_1, p_2, \dots, p_m$ and is strictly greater than $1$, so there must exist a new prime $p_{m+1}\in S$.