Note that if $P(n) = 0$ then $P(P(n) + n) = P(n) = 0$ which is not prime. Let $P(x)$ be a degree $k$ polynomial of the form $P(x) = a_k x^k + a_{k-1} x^{k-1} + \dots + a_0$ and note that if $P(n) \ne 0$ then
$$P(P(n) + n) - P(n) = a_k[(P(n) + n)^k - n^k] + a_{k-1}[(P(n) + n)^{k-1} - n^{k-1}] + \dots + a_1 P(n)$$
which is divisible by $(P(n)+n) - n = P(n)$. Therefore if $P(P(n)+n)$ is prime then either $P(n) = \pm 1$ or $P(P(n)+n) = \pm P(n) = p$ for some prime number $p$. Since $P(x)$ is a polynomial, it follows that $P(n) = \pm 1$ for only finitely many integers $n$. Therefore either $P(n) = P(P(n)+n)$ for infinitely many integers $n$ or $P(n) = -P(P(n)+n)$ for infinitely many integers $n$.

Suppose that $P(n) = P(P(n)+n)$ for infinitely many integers $n$. This implies that the polynomial $P(P(x)+x) - P(x)$ has infinitely many roots and thus is identically zero. Therefore $P(P(x)+x) = P(x)$ holds identically. Now note that if $k \ge 2$ then $P(P(x)+x)$ has degree $k^2$ while $P(x)$ has degree $k$, which is not possible. Therefore $P(x)$ is at most linear with $P(x) = ax + b$ for some integers $a$ and $b$. Now note that
$$P(P(x) + x) = a(a + 1)x + ab + b$$
and thus $a = a(a + 1)$ and $ab + b = b$. It follows that $a = 0$ which leads to the solution $P(n) = p$ where $p$ is a prime number.

By the same argument if $P(n) = -P(P(n)+n)$ for infinitely many integers $n$ then $P(x) = -P(P(x)+x)$ holds identically and $P(x)$ is linear with $P(x) = ax + b$. In this case it follows that $a = -a(a + 1)$ and $ab + b = -b$. This implies that either $a = 0$ or $a = -2$. If $a = -2$ then $P(n) = -2n + b$ which is prime for some integers $n$ only if $b$ is odd. Note that in this case $P(P(n)+n) = 2n - b$ which is indeed prime for infinitely many integers $n$ as long as $b$ is odd. $\square$