Assume that there are only finitely many primes $\equiv 1 \mod p$ and let $P$ be their product ($P=1$ if there are none). Let $q$ be any prime divisor of $(P+1)^p-P^p$. We have $q$ coprime to $P$. Set $Q=\frac{P+1}{P} \mod q$, then $Q^p = \left( \frac{P+1}{P} \right)^p = 1 \mod q$. Now $Q$ has order $1$ or $p$ seen $\mod q$, and order $1$ is impossible, so it has order $p$. By this, $p|q-1$, thus $q \equiv 1 \mod p$, contradiction (we found such a prime definitely not dividing $P$). Since $p$ is odd, every prime of type $pn+1$ is of type $2pm+1$.

Assume $ q_i, 1\leq i \leq n$ are the only such primes. Let $ P=q_1q_2 \dots q_n$, $ N = P^p - 1$. Let q be a prime divisor of N. $ d = \text{ord}_q(P)|p \implies d=p$, by FLT p|q-1 and since p is odd, $ q=2kp+1$. Obviously $ q>q_i$. Repeating the argument there is another such prime and hence infinitely many.

KevinB wrote: Assume $ q_i, 1\leq i \leq n$ are the only such primes. Let $ P = q_1q_2 \dots q_n$, $ N = P^p - 1$. Let q be a prime divisor of N. $ d = \text{ord}_q(P)|p \implies d = p$, by FLT p|q-1 and since p is odd, $ q = 2kp + 1$. Obviously $ q > q_i$. Repeating the argument there is another such prime and hence infinitely many. But for FLT you need to show that $ q \neq 2$ and that $ d \neq 1$. I think it's better to look at $ (P-1)^p - 1$ instead, since it's odd and also not divisible by $ p$. For your proof you also need $ P \neq 1$, but that can be shown by looking at $ 2^p - 1$ for example.