Find all polynomials $P(x)$ with integer coefficients such that $P(P(n) + n)$ is a prime number for infinitely many integers $n$.
1. **Lemma:** For integer coefficient polynomials, \(a - b \mid P(a) - P(b)\), where \(a\) and \(b\) are integers.

   **Proof:** We have 
   \[
   P(a) - P(b) = c_y a^y + c_{y-1} a^{y-1} + \dots + c_1 a + c_0 - (c_y b^y + c_{y-1} b^{y-1} + \dots + c_1 b + c_0)
   \]
   \[
   = c_y (a^y - b^y) + c_{y-1} (a^{y-1} - b^{y-1}) + \dots + c_1 (a - b)
   \]
   Note that \(a - b \mid a^k - b^k = (a - b)(a^{k-1} + a^{k-2}b + \dots + b^{k-1})\) for all \(k \geq 2\). This implies \(a - b \mid P(a) - P(b)\), since every term of this expression is a multiple of \(a - b\).

2. **Claim:** \(P(n) \mid P(P(n) + n)\).

   **Proof:** Plugging in \(a = P(n) + n\) and \(b = n\) in our lemma, we have 
   \[
   a - b \mid P(a) - P(b) \implies P(n) + n - n \mid P(P(n) + n) - P(n) \implies P(n) \mid P(P(n) + n) - P(n)
   \]
   Since \(P(n) \mid P(n)\), it follows that \(P(n) \mid P(P(n) + n)\).

3. Since \(P(P(n) + n)\) is prime for infinitely many \(n\), this means we must have \(P(n) = \pm 1\) or \(P(n) = \pm P(P(n) + n)\) for infinitely many \(n\).

4. **Case 1:** \(P(n) = \pm 1\) for infinitely many \(n\).

   Since \(P\) is a polynomial, and there are infinitely many values of \(n\) for which \(P(n) = \pm 1\), it follows that \(P(x) = 1\) or \(P(x) = -1\). Clearly, \(1\) and \(-1\) are not primes, so this case is impossible.

5. **Case 2:** \(P(n) = \pm P(P(n) + n)\) for infinitely many \(n\).

   Since \(P\) is a polynomial, by the same logic as above, it follows that \(P(n) = \pm P(P(n) + n)\) for all values of \(n\). This means we have \(\deg(P(n)) = \deg(\pm P(P(n) + n))\). As defined at the start, \(\deg(P) = y\). The highest power, or degree of \(P(P(n) + n)\) will be \((n^y)^y = n^{y^2}\), so we have \(y = y^2\), so \(y = 0\) or \(y = 1\).

6. The case \(y = 0\), so when \(P\) is a constant, is obvious; it's just \(P(x) = p\) where \(p\) is a prime.

7. Now we will look at when \(y = 1\). This means \(P\) is linear, and we can rewrite \(P(x) = mx + k\). Plugging this in our equality \(P(n) = \pm P(P(n) + n)\) and matching coefficients, we obtain that \(m = 0\) or \(m = -2\). 

   - \(m = 0\) yields the constant solution we found earlier.
   - \(m = -2\) means \(P(n) = -2n + k\). If we want this to be prime for some values of \(n\), then \(k\) must be odd.

Therefore, our solutions are:
\[
\boxed{P(x) = p \text{ where } p \text{ is a prime, and } P(x) = -2x + k \text{ where } k \text{ is an odd integer.}}
\]