Find all integer triples $(m,p,q)$ satisfying \[2^mp^2+1=q^5\] where $m>0$ and both $p$ and $q$ are prime numbers.
We are given the equation:
\[2^m p^2 + 1 = q^5\]
where \(m > 0\) and both \(p\) and \(q\) are prime numbers. We need to find all integer triples \((m, p, q)\) that satisfy this equation.

1. **Rearrange and Factorize:**
   \[2^m p^2 = q^5 - 1\]
   We can factorize the right-hand side using the difference of powers:
   \[q^5 - 1 = (q - 1) \left(\frac{q^5 - 1}{q - 1}\right)\]
   Therefore, we have:
   \[2^m p^2 = (q - 1) \left(\frac{q^5 - 1}{q - 1}\right)\]

2. **GCD Analysis:**
   Note that \(\gcd(q - 1, \frac{q^5 - 1}{q - 1}) \mid 5\). This is because \(\frac{q^5 - 1}{q - 1}\) is a polynomial in \(q\) with integer coefficients, and the only prime factors of \(q^5 - 1\) that are not factors of \(q - 1\) are those dividing 5.

3. **Case 1: \(\gcd(q - 1, \frac{q^5 - 1}{q - 1}) = 1\)**

   - **Subcase 1.1: \(p = 2\)**
     \[2^{m+2} = (q - 1) \left(\frac{q^5 - 1}{q - 1}\right)\]
     Since \(q^5 - 1 > (q - 1)^2\), we must have \(q - 1 = 1 \Rightarrow q = 2\). This gives:
     \[\frac{2^5 - 1}{2 - 1} = 31\]
     However, \(2^{m+2} = 31\) has no integer solutions for \(m\).

   - **Subcase 1.2: \(p \neq 2\)**
     We have two possibilities:
     - \(q - 1 = 2^m\) and \(\frac{q^5 - 1}{q - 1} = p^2\)
     - \(q - 1 = p^2\) and \(\frac{q^5 - 1}{q - 1} = 2^m\)

     For the first possibility:
     \[q - 1 = 2^m \Rightarrow q = 2^m + 1\]
     Substituting back, we get:
     \[2^m p^2 = (2^m + 1)^5 - 1\]
     Simplifying, we get:
     \[p^2 = \frac{(2^m + 1)^5 - 1}{2^m}\]

     For the second possibility:
     \[q - 1 = p^2 \Rightarrow q = p^2 + 1\]
     Substituting back, we get:
     \[2^m = \frac{(p^2 + 1)^5 - 1}{p^2}\]

     - If \(m\) is odd:
       \[q = 2^m + 1\]
       For \(m = 1\):
       \[q = 2 + 1 = 3\]
       \[2^1 p^2 + 1 = 3^5 \Rightarrow 2p^2 + 1 = 243 \Rightarrow 2p^2 = 242 \Rightarrow p^2 = 121 \Rightarrow p = 11\]
       This gives the solution \((m, p, q) = (1, 11, 3)\).

     - If \(m\) is even:
       \[q = 4^k + 1\]
       For \(k \geq 2\):
       \[p^2 = \frac{(4^k + 1)^5 - 1}{4^k}\]
       This yields a contradiction since \(p^2 \equiv 5 \pmod{8}\) is not possible.

4. **Case 2: \(\gcd(q - 1, \frac{q^5 - 1}{q - 1}) = 5\)**
   Clearly, we must have \(p = 5\):
   \[q - 1 = 5 \Rightarrow q = 6\]
   This is a contradiction since \(q\) must be a prime number.

We conclude from our search that \(\boxed{(m, p, q) = (1, 11, 3)}\) is the only solution.