Find all triples $(p,M, z)$ of integers, where $p$ is prime, $m$ is positive and $z$ is negative, that satisfy the equation
$$p^3 + pm + 2zm = m^2 + pz + z^2$$
1. Given the equation:
   \[
   p^3 + pm + 2zm = m^2 + pz + z^2
   \]
   where \( p \) is a prime number, \( m \) is a positive integer, and \( z \) is a negative integer.

2. Let \( x = m - z \). Then, we can rewrite \( m \) as \( m = x + z \).

3. Substitute \( m = x + z \) into the original equation:
   \[
   p^3 + p(x + z) + 2z(x + z) = (x + z)^2 + pz + z^2
   \]

4. Simplify the equation:
   \[
   p^3 + px + pz + 2zx + 2z^2 = x^2 + 2xz + z^2 + pz + z^2
   \]

5. Combine like terms:
   \[
   p^3 + px + pz + 2zx + 2z^2 = x^2 + 2xz + z^2 + pz + z^2
   \]

6. Cancel out the common terms on both sides:
   \[
   p^3 + px + 2z^2 = x^2 + z^2
   \]

7. Rearrange the equation:
   \[
   p^3 + px + z^2 = x^2
   \]

8. Notice that \( p \mid x \) because \( p \) is a prime number. Let \( x = pa \), where \( a \) is an integer.

9. Substitute \( x = pa \) into the equation:
   \[
   p^3 + p(pa) + z^2 = (pa)^2
   \]

10. Simplify the equation:
    \[
    p^3 + p^2a + z^2 = p^2a^2
    \]

11. Divide the entire equation by \( p^2 \):
    \[
    \frac{p^3}{p^2} + \frac{p^2a}{p^2} + \frac{z^2}{p^2} = a^2
    \]

12. Simplify:
    \[
    p + a + \frac{z^2}{p^2} = a^2
    \]

13. Since \( z \) is a negative integer, \( z^2 \) is positive. For the equation to hold, \( \frac{z^2}{p^2} \) must be an integer. This implies \( p^2 \mid z^2 \), and since \( p \) is prime, \( p \mid z \).

14. Let \( z = -pb \), where \( b \) is a positive integer. Substitute \( z = -pb \) into the equation:
    \[
    p + a + \frac{(-pb)^2}{p^2} = a^2
    \]

15. Simplify:
    \[
    p + a + b^2 = a^2
    \]

16. Rearrange the equation:
    \[
    a^2 - a - p - b^2 = 0
    \]

17. Solve the quadratic equation for \( a \):
    \[
    a = \frac{1 \pm \sqrt{1 + 4(p + b^2)}}{2}
    \]

18. For \( a \) to be an integer, \( 1 + 4(p + b^2) \) must be a perfect square. Let \( k^2 = 1 + 4(p + b^2) \), where \( k \) is an integer.

19. Solve for \( p \):
    \[
    p + b^2 = \frac{k^2 - 1}{4}
    \]

20. Since \( p \) is a prime number, \( \frac{k^2 - 1}{4} - b^2 \) must be a prime number. Check possible values of \( k \) and \( b \) to find valid solutions.

21. For \( a = 2 \), we have:
    \[
    2^2 - 2 - p - b^2 = 0 \implies 4 - 2 - p - b^2 = 0 \implies 2 = p + b^2
    \]

22. Since \( p \) is prime, the only possible value is \( p = 2 \) and \( b = 0 \). However, \( b \) must be positive, so this case is not valid.

23. For \( a = 2 \) and \( p = 2 \), we have:
    \[
    x = pa = 2 \cdot 2 = 4
    \]

24. Therefore, \( m = z + 4 \). Since \( m > 0 \) and \( z < 0 \), we have \( z \in \{-1, -2, -3\} \).

25. The solutions are:
    \[
    (2, 4 + z, z) \text{ for } z \in \{-1, -2, -3\}
    \]

The final answer is \( \boxed{ (2, 4 + z, z) \text{ for } z \in \{-1, -2, -3\} } \).