Find all integers $n$ such that $2n+1$ is a divisor of $(n!)^2 + 2^n$.

[i]Proposed by Silouanos Brazitikos[/i]
To find all integers \( n \) such that \( 2n+1 \) is a divisor of \( (n!)^2 + 2^n \), we can proceed as follows:

1. **Express the problem in terms of \( p \):**
   Let \( p = 2n + 1 \). Then \( n = \frac{p-1}{2} \). We need to find \( p \) such that \( p \) divides \( (n!)^2 + 2^n \).

2. **Use Wilson's Theorem:**
   Wilson's Theorem states that for a prime \( p \), \((p-1)! \equiv -1 \pmod{p}\). Since \( p = 2n + 1 \), we have:
   \[
   (2n)! \equiv -1 \pmod{p}
   \]
   We need to relate this to \( (n!)^2 \).

3. **Relate \( (n!)^2 \) to \( (2n)! \):**
   Using the fact that \( (2n)! = (2n)(2n-1)\cdots(n+1) \cdot n! \), we can write:
   \[
   (2n)! = (2n)(2n-1)\cdots(n+1) \cdot n!
   \]
   Since \( n = \frac{p-1}{2} \), we have:
   \[
   (2n)! = (p-1)! \equiv -1 \pmod{p}
   \]
   Therefore,
   \[
   (n!)^2 \equiv (-1)^n (2n)! \pmod{p}
   \]

4. **Simplify the congruence:**
   Since \( (2n)! \equiv -1 \pmod{p} \), we have:
   \[
   (n!)^2 \equiv (-1)^n (-1) \pmod{p}
   \]
   \[
   (n!)^2 \equiv (-1)^{n+1} \pmod{p}
   \]

5. **Consider the term \( 2^n \):**
   We need \( p \) to divide \( (n!)^2 + 2^n \). Thus,
   \[
   (n!)^2 + 2^n \equiv 0 \pmod{p}
   \]
   Substituting the congruence for \( (n!)^2 \):
   \[
   (-1)^{n+1} + 2^n \equiv 0 \pmod{p}
   \]
   \[
   2^n \equiv (-1)^{n+1} \pmod{p}
   \]

6. **Analyze the powers of \(-1\):**
   We need to determine when \( 2^n \equiv (-1)^{n+1} \pmod{p} \). This depends on the parity of \( n \):
   - If \( n \) is even, \( (-1)^{n+1} = -1 \).
   - If \( n \) is odd, \( (-1)^{n+1} = 1 \).

7. **Use quadratic reciprocity:**
   We use the fact that \( 2^{\frac{p-1}{2}} \equiv \left( \frac{2}{p} \right) \pmod{p} \), where \( \left( \frac{2}{p} \right) \) is the Legendre symbol. By quadratic reciprocity:
   \[
   \left( \frac{2}{p} \right) = (-1)^{\frac{p^2-1}{8}}
   \]
   Thus,
   \[
   2^{\frac{p-1}{2}} \equiv (-1)^{\frac{p^2-1}{8}} \pmod{p}
   \]

8. **Determine \( p \mod 8 \):**
   We need \( (-1)^{n+1} \equiv 2^n \pmod{p} \). This implies that \( p \equiv 1 \) or \( 3 \pmod{8} \).

Conclusion:
The integers \( n \) such that \( 2n+1 \) is a divisor of \( (n!)^2 + 2^n \) are given by:
\[
n = \frac{p-1}{2} \quad \text{where} \quad p \equiv 1 \text{ or } 3 \pmod{8}
\]

The final answer is \( \boxed{ n = \frac{p-1}{2} \quad \text{where} \quad p \equiv 1 \text{ or } 3 \pmod{8} } \)