Find all pairs of primes $(p, q)$ such that $$p^3 - q^5 = (p + q)^2.$$
1. We start with the given equation:
   \[
   p^3 - q^5 = (p + q)^2
   \]
   where \( p \) and \( q \) are prime numbers.

2. Let's first test some small prime values for \( p \) and \( q \).

3. Let \( p = 3 \):
   \[
   3^3 - q^5 = (3 + q)^2
   \]
   Simplifying, we get:
   \[
   27 - q^5 = 9 + 6q + q^2
   \]
   Rearranging terms, we obtain:
   \[
   q^5 + q^2 + 6q - 18 = 0
   \]
   Using the Rational Root Theorem (IRT), we test possible rational roots, but find no integer solutions for \( q \).

4. Let \( q = 3 \):
   \[
   p^3 - 3^5 = (p + 3)^2
   \]
   Simplifying, we get:
   \[
   p^3 - 243 = p^2 + 6p + 9
   \]
   Rearranging terms, we obtain:
   \[
   p^3 - p^2 - 6p - 252 = 0
   \]
   Using the Rational Root Theorem (IRT), we test possible rational roots and find that \( p = 7 \) is a solution. Substituting \( p = 7 \) back into the equation confirms it:
   \[
   7^3 - 3^5 = (7 + 3)^2
   \]
   \[
   343 - 243 = 100
   \]
   \[
   100 = 100
   \]
   Thus, \((p, q) = (7, 3)\) is a solution.

5. Now, we consider the congruence properties of \( p \) and \( q \) modulo 3. We assume \( p, q \equiv 1 \pmod{3} \):
   \[
   p^3 \equiv 1^3 \equiv 1 \pmod{3}
   \]
   \[
   q^5 \equiv 1^5 \equiv 1 \pmod{3}
   \]
   \[
   (p + q)^2 \equiv (1 + 1)^2 \equiv 2^2 \equiv 4 \equiv 1 \pmod{3}
   \]
   This gives:
   \[
   1 - 1 \equiv 1 \pmod{3}
   \]
   which is a contradiction.

6. Similarly, for the cases \( p \equiv 1 \pmod{3} \) and \( q \equiv 2 \pmod{3} \) or \( p \equiv 2 \pmod{3} \) and \( q \equiv 1 \pmod{3} \):
   \[
   p^3 \equiv 1 \pmod{3}, \quad q^5 \equiv 2 \pmod{3}
   \]
   \[
   (p + q)^2 \equiv (1 + 2)^2 \equiv 3^2 \equiv 0 \pmod{3}
   \]
   This gives:
   \[
   1 - 2 \equiv 0 \pmod{3}
   \]
   which is a contradiction.

7. Finally, for the case \( p, q \equiv 2 \pmod{3} \):
   \[
   p^3 \equiv 2^3 \equiv 8 \equiv 2 \pmod{3}
   \]
   \[
   q^5 \equiv 2^5 \equiv 32 \equiv 2 \pmod{3}
   \]
   \[
   (p + q)^2 \equiv (2 + 2)^2 \equiv 4^2 \equiv 1 \pmod{3}
   \]
   This gives:
   \[
   2 - 2 \equiv 1 \pmod{3}
   \]
   which is a contradiction.

Thus, the only solution pair is \((p, q) = (7, 3)\).

The final answer is \( \boxed{ (p, q) = (7, 3) } \).