Determine the integers $x$ such that $2^x + x^2 + 25$ is the cube of a prime number
1. Let \( 2^x + x^2 + 25 = p^3 \) where \( p \) is a prime number. Since \( p^3 > 25 \) and \( p^3 \) must be greater than \( 2^x \) and \( x^2 \), we know \( p \) must be greater than 2. Therefore, \( p \) is an odd prime number.

2. If \( x < 0 \), then \( 2^x \not\in \mathbb{Z} \) (since \( 2^x \) would be a fraction), so we must have \( x \in \mathbb{N} \).

3. Check \( x = 0 \):
   \[
   2^0 + 0^2 + 25 = 1 + 0 + 25 = 26
   \]
   Since 26 is not a cube of a prime number, \( x \) must be greater than 0.

4. Since \( p \) is odd, \( p^3 \) is odd. Therefore, \( 2^x + x^2 + 25 \) must be odd. This implies \( x \) must be even (since \( 2^x \) is even for even \( x \), and \( x^2 \) is even for even \( x \)).

5. Let \( x = 2k \) for some integer \( k \). Then:
   \[
   2^{2k} + (2k)^2 + 25 = p^3
   \]

6. Check \( p = 3 \):
   \[
   3^3 = 27
   \]
   \[
   2^{2k} + 4k^2 + 25 = 27
   \]
   \[
   2^{2k} + 4k^2 = 2
   \]
   This is not possible since \( 2^{2k} \geq 4 \) for \( k \geq 1 \).

7. Therefore, \( p > 3 \). Since \( p \) is a prime number and \( p^3 \equiv 1 \pmod{3} \), we need:
   \[
   2^{2k} + 4k^2 + 25 \equiv 1 \pmod{3}
   \]
   \[
   2^{2k} \equiv 1 \pmod{3} \quad \text{(since \( 2^2 \equiv 1 \pmod{3} \))}
   \]
   \[
   4k^2 \equiv 1 \pmod{3}
   \]
   This implies \( k \equiv 0 \pmod{3} \). Let \( k = 3a \).

8. Substitute \( k = 3a \):
   \[
   x = 2k = 6a
   \]
   \[
   2^{6a} + 36a^2 + 25 = p^3
   \]

9. Since \( p^3 > 2^{6a} \), we have \( p \geq 2^{2a} + 1 \). Therefore:
   \[
   2^{6a} + 36a^2 + 25 \geq (2^{2a} + 1)^3
   \]
   \[
   2^{6a} + 36a^2 + 25 \geq 2^{6a} + 3 \cdot 2^{4a} + 3 \cdot 2^{2a} + 1
   \]
   \[
   36a^2 + 25 \geq 3 \cdot 2^{4a} + 3 \cdot 2^{2a} + 1
   \]

10. For \( a > 1 \), \( 3 \cdot 2^{4a} + 3 \cdot 2^{2a} + 1 \) grows faster than \( 36a^2 + 25 \). Therefore, we need to check small values of \( a \).

11. Check \( a = 1 \):
    \[
    x = 6 \cdot 1 = 6
    \]
    \[
    2^6 + 6^2 + 25 = 64 + 36 + 25 = 125 = 5^3
    \]
    Since 5 is a prime number, \( x = 6 \) is a solution.

12. Verify if there are other solutions:
    For \( a > 1 \), the inequality \( 36a^2 + 25 \geq 3 \cdot 2^{4a} + 3 \cdot 2^{2a} + 1 \) does not hold. Therefore, there are no other solutions.

Conclusion:
The only solution is \( x = 6 \).

The final answer is \( \boxed{ x = 6 } \).