For how many pairs of primes $(p, q)$, is $p^2 + 2pq^2 + 1$ also a prime?
1. **Identify the form of the expression:**
   We need to determine for how many pairs of primes \((p, q)\), the expression \(p^2 + 2pq^2 + 1\) is also a prime.

2. **Check the parity of the expression:**
   Notice that \(p^2 + 2pq^2 + 1\) must be odd for it to be a prime (since the only even prime is 2, and the expression is greater than 2 for any prime \(p\) and \(q\)). Therefore, \(p\) must be an odd prime.

3. **Consider \(p = 2\):**
   If \(p = 2\), then the expression becomes:
   \[
   2^2 + 2 \cdot 2 \cdot q^2 + 1 = 4 + 4q^2 + 1 = 4q^2 + 5
   \]
   For \(4q^2 + 5\) to be a prime, \(q\) must be such that \(4q^2 + 5\) is a prime number.

4. **Check small values of \(q\):**
   - For \(q = 2\):
     \[
     4 \cdot 2^2 + 5 = 4 \cdot 4 + 5 = 16 + 5 = 21 \quad (\text{not a prime})
     \]
   - For \(q = 3\):
     \[
     4 \cdot 3^2 + 5 = 4 \cdot 9 + 5 = 36 + 5 = 41 \quad (\text{a prime})
     \]

5. **Consider \(p \geq 5\):**
   All primes \(p \geq 5\) are of the form \(6k \pm 1\). We need to check if \(p^2 + 2pq^2 + 1\) can be a prime for these values.

6. **Modulo 3 analysis:**
   For \(p = 6k \pm 1\), we have:
   \[
   p^2 \equiv 1 \pmod{3}
   \]
   and
   \[
   2pq^2 \equiv 2 \cdot (6k \pm 1) \cdot q^2 \equiv 2q^2 \pmod{3}
   \]
   Therefore,
   \[
   p^2 + 2pq^2 + 1 \equiv 1 + 2q^2 + 1 \equiv 2 + 2q^2 \pmod{3}
   \]
   For \(p^2 + 2pq^2 + 1\) to be a prime, it must not be divisible by 3. Hence,
   \[
   2 + 2q^2 \not\equiv 0 \pmod{3}
   \]
   This implies:
   \[
   2q^2 \not\equiv 1 \pmod{3}
   \]
   Since \(q\) is a prime, \(q\) can be 2 or 3 modulo 3. If \(q \equiv 1 \pmod{3}\), then \(q^2 \equiv 1 \pmod{3}\), and \(2q^2 \equiv 2 \pmod{3}\), which is not 1 modulo 3. Therefore, \(q\) must be 2 or 3.

7. **Check \(q = 2\) and \(q = 3\) for \(p \geq 5\):**
   - For \(q = 2\):
     \[
     p^2 + 2p \cdot 2^2 + 1 = p^2 + 8p + 1
     \]
     This expression is not necessarily a prime for any \(p \geq 5\).
   - For \(q = 3\):
     \[
     p^2 + 2p \cdot 3^2 + 1 = p^2 + 18p + 1
     \]
     This expression is not necessarily a prime for any \(p \geq 5\).

8. **Conclusion:**
   The only pair \((p, q)\) that works is \((2, 3)\).

The final answer is \( \boxed{ (2, 3) } \)