Find all pairs of prime numbers $(p, q)$ for which $7pq^2 + p = q^3 + 43p^3 + 1$
To find all pairs of prime numbers \((p, q)\) for which the equation \(7pq^2 + p = q^3 + 43p^3 + 1\) holds, we will analyze the equation under different cases.

1. **Case 1: Both \(p\) and \(q\) are odd primes**

   If both \(p\) and \(q\) are odd, then:
   - The left-hand side (LHS) \(7pq^2 + p\) is even because \(7pq^2\) is odd (product of odd numbers) and \(p\) is odd, making the sum even.
   - The right-hand side (RHS) \(q^3 + 43p^3 + 1\) is odd because \(q^3\) and \(43p^3\) are odd (odd numbers raised to an odd power remain odd), and adding 1 to an even number results in an odd number.

   Since the LHS is even and the RHS is odd, this case is impossible. Therefore, at least one of \(p\) or \(q\) must be 2.

2. **Case 2: \(p = 2\)**

   Substitute \(p = 2\) into the equation:
   \[
   7 \cdot 2 \cdot q^2 + 2 = q^3 + 43 \cdot 2^3 + 1
   \]
   Simplify:
   \[
   14q^2 + 2 = q^3 + 344 + 1
   \]
   \[
   14q^2 + 2 = q^3 + 345
   \]
   Rearrange the equation:
   \[
   q^3 - 14q^2 = -343
   \]
   \[
   q^3 - 14q^2 + 343 = 0
   \]
   Factorize the equation:
   \[
   q^2(q - 14) = 343
   \]
   Since \(343 = 7^3\), we can write:
   \[
   q^2(14 - q) = 7^3
   \]
   The only integer solution for \(q\) that satisfies this equation is \(q = 7\).

   Therefore, one solution is \((p, q) = (2, 7)\).

3. **Case 3: \(q = 2\)**

   Substitute \(q = 2\) into the equation:
   \[
   7p \cdot 2^2 + p = 2^3 + 43p^3 + 1
   \]
   Simplify:
   \[
   28p + p = 8 + 43p^3 + 1
   \]
   \[
   29p = 43p^3 + 9
   \]
   Rearrange the equation:
   \[
   43p^3 - 29p + 9 = 0
   \]
   Factor out \(p\):
   \[
   p(43p^2 - 29) = -9
   \]
   Since \(p\) is a prime number, \(p\) must be a factor of 9. The only prime factor of 9 is 3, but substituting \(p = 3\) does not satisfy the equation. Therefore, there are no solutions in this case.

Therefore, the only solution is \((p, q) = (2, 7)\).

The final answer is \( \boxed{ (p, q) = (2, 7) } \).