Find all positive integers $m,n$ and prime numbers $p$ for which $\frac{5^m+2^np}{5^m-2^np}$ is a perfect square.
To find all positive integers \( m, n \) and prime numbers \( p \) for which \(\frac{5^m + 2^n p}{5^m - 2^n p}\) is a perfect square, we start by analyzing the given expression.

1. **GCD Analysis**:
   \[
   \gcd(5^m + 2^n p, 5^m - 2^n p) = \gcd(5^m + 2^n p, 2 \cdot 5^m)
   \]
   This is because:
   \[
   \gcd(a, b) = \gcd(a, b - a)
   \]
   Applying this property:
   \[
   \gcd(5^m + 2^n p, 5^m - 2^n p) = \gcd(5^m + 2^n p, 2 \cdot 2^n p) = \gcd(5^m + 2^n p, 2^{n+1} p)
   \]
   Hence, their gcd divides:
   \[
   \gcd(2 \cdot 5^m, 2^{n+1} p) = 2, 10
   \]
   Therefore, the gcd must be one of \(1, 2, 5, 10\).

2. **Divisibility Condition**:
   For \(\frac{5^m + 2^n p}{5^m - 2^n p}\) to be an integer, \(5^m - 2^n p\) must divide the gcd. Thus:
   \[
   5^m - 2^n p = 1, 2, 5, 10
   \]
   We can eliminate \(2\) and \(10\) because \(m, n \ge 1\) and considering parity, we get a contradiction.

3. **Case Analysis**:
   - **Case 1: \(5^m - 2^n p = 1\)**:
     \[
     5^m = 2^n p + 1
     \]
     Observe that:
     \[
     2^{n+1} p = k^2 - 1 = (k-1)(k+1)
     \]
     Since \(k\) is odd, \(k-1\) and \(k+1\) are even. We have two subcases:
     - \(k-1 = 2\) and \(k+1 = 2^n p\):
       \[
       k = 3 \implies 2^n p = 8p \implies p = 2, n = 1, m = 1
       \]
     - \(k-1 = 2p\) and \(k+1 = 2^n\):
       \[
       2p = 2^{n-1} - 1 \quad \text{or} \quad 2p = 2^{n-1} + 1
       \]
       - If \(2^{n-1} - 1 = p\):
         \[
         5^m - 2^n (2^{n-1} - 1) = 1 \implies 5^m - 2^{2n-1} + 2^n = 1
         \]
         This implies:
         \[
         5^m \equiv 1 \pmod{2^n} \implies 2^{n-2} \mid m
         \]
         For \(n \ge 3\), \(5^m - 2^{2n-1} + 2^n > 1\), contradiction.
       - If \(2^{n-1} + 1 = p\):
         \[
         5^m - 2^{2n-1} - 2^n = 1
         \]
         Similarly:
         \[
         2^{n-2} \mid m
         \]
         For \(n \ge 4\), \(5^m - 2^{2n-1} - 2^n > 1\), contradiction. For \(n = 3\):
         \[
         n = 3, p = 3, m = 2
         \]

   - **Case 2: \(5^m - 2^n p = 5\)**:
     \[
     5^m = 2^n p + 5
     \]
     Clearly, \(p = 5\):
     \[
     5^{m-1} = 2^n + 1
     \]
     By the Catalan conjecture (Mihăilescu's theorem), the only solution is:
     \[
     n = 2, m = 2, p = 5
     \]

Hence, the only solutions are:
\[
(n, m, p) = (1, 1, 2), (3, 2, 3), (2, 2, 5)
\]

The final answer is \( \boxed{ (n, m, p) = (1, 1, 2), (3, 2, 3), (2, 2, 5) } \).