Find all triples $(m,p,q)$ where $ m $ is a positive integer and $ p , q $ are primes.
\[ 2^m p^2 + 1 = q^5 \]
We are given the equation:
\[ 2^m p^2 + 1 = q^5 \]
where \( m \) is a positive integer and \( p \) and \( q \) are primes. We need to find all possible triples \((m, p, q)\).

1. **Rewrite the equation:**
   \[ 2^m p^2 = q^5 - 1 \]
   Notice that \( q^5 - 1 \) can be factored using the difference of powers:
   \[ q^5 - 1 = (q-1)(q^4 + q^3 + q^2 + q + 1) \]
   Therefore, we have:
   \[ 2^m p^2 = (q-1)(q^4 + q^3 + q^2 + q + 1) \]

2. **Analyze the factors:**
   Since \( p \) and \( q \) are primes, and \( 2^m p^2 \) is a product of powers of 2 and a prime squared, we need to consider the possible values of \( q \).

3. **Case 1: \( q = 2 \)**
   If \( q = 2 \), then:
   \[ 2^m p^2 = (2-1)(2^4 + 2^3 + 2^2 + 2 + 1) = 1 \cdot 31 = 31 \]
   This implies:
   \[ 2^m p^2 = 31 \]
   Since 31 is a prime number and not a power of 2 times a square of a prime, this case fails.

4. **Case 2: \( q \neq 2 \)**
   Since \( q \) is an odd prime, \( q-1 \) is even. Let \( q-1 = 2^a \cdot b \) where \( b \) is odd. Then:
   \[ 2^m p^2 = 2^a \cdot b \cdot (q^4 + q^3 + q^2 + q + 1) \]
   Since \( 2^m p^2 \) is a product of powers of 2 and a prime squared, we need to consider the greatest common divisor (gcd) of \( q-1 \) and \( q^4 + q^3 + q^2 + q + 1 \).

5. **GCD Analysis:**
   \[ \gcd(q-1, q^4 + q^3 + q^2 + q + 1) \]
   Since \( q^4 + q^3 + q^2 + q + 1 \equiv 5 \pmod{q-1} \), the gcd can be either 1 or 5.

6. **Subcase 1: \( \gcd(q-1, q^4 + q^3 + q^2 + q + 1) = 1 \)**
   This implies \( q-1 = 2^m \) and \( q^4 + q^3 + q^2 + q + 1 = p^2 \).

   - If \( q = 3 \):
     \[ q^4 + q^3 + q^2 + q + 1 = 3^4 + 3^3 + 3^2 + 3 + 1 = 81 + 27 + 9 + 3 + 1 = 121 = 11^2 \]
     Thus, \( p = 11 \) and \( q = 3 \), and \( q-1 = 2 \), so \( m = 1 \).
     Therefore, we have the solution:
     \[ (m, p, q) = (1, 11, 3) \]

   - For \( q > 3 \):
     \[ (2q^2 + q)^2 < 4p^2 < (2q^2 + q + 1)^2 \]
     This inequality shows that \( q^4 + q^3 + q^2 + q + 1 \) cannot be a perfect square for \( q > 3 \).

7. **Subcase 2: \( \gcd(q-1, q^4 + q^3 + q^2 + q + 1) = 5 \)**
   This implies \( p = 5 \) and \( q-1 = 2^a \cdot 5 \).

   - Since \( q^4 + q^3 + q^2 + q + 1 \) is odd, it must be equal to 5, which is a contradiction because \( q^4 + q^3 + q^2 + q + 1 \) is much larger than 5 for any prime \( q \).

Therefore, the only solution is:
\[ (m, p, q) = (1, 11, 3) \]

The final answer is \( \boxed{ (1, 11, 3) } \)