Find all positive integer $n$ and prime number $p$ such that $p^2+7^n$ is a perfect square 
1. **Initial Setup:**
   We start with the equation \( p^2 + 7^n = k^2 \) where \( p \) is a prime number, \( n \) is a positive integer, and \( k \) is an integer. We need to find all pairs \((n, p)\) such that this equation holds.

2. **Modulo 3 Analysis:**
   Notice that for \( p \geq 5 \) and \( p = 2 \), we have:
   \[
   p^2 \equiv 1 \pmod{3} \quad \text{or} \quad p^2 \equiv 4 \equiv 1 \pmod{3}
   \]
   Therefore,
   \[
   p^2 + 7^n \equiv 1 + 7^n \pmod{3}
   \]
   Since \( 7 \equiv 1 \pmod{3} \), we have:
   \[
   7^n \equiv 1^n \equiv 1 \pmod{3}
   \]
   Thus,
   \[
   p^2 + 7^n \equiv 1 + 1 \equiv 2 \pmod{3}
   \]
   However, \( k^2 \equiv 0 \) or \( 1 \pmod{3} \). This is a contradiction because \( 2 \not\equiv 0 \) or \( 1 \pmod{3} \). Therefore, \( p \neq 2 \) and \( p \geq 5 \) is not possible. Hence, \( p = 3 \).

3. **Case \( p = 3 \):**
   Substituting \( p = 3 \) into the equation, we get:
   \[
   9 + 7^n = k^2
   \]
   Rearranging, we have:
   \[
   7^n = k^2 - 9
   \]
   This can be written as:
   \[
   7^n = (k - 3)(k + 3)
   \]

4. **Prime Factorization Analysis:**
   Since \( 7 \) is a prime number, the factors \( (k - 3) \) and \( (k + 3) \) must be powers of \( 7 \). Let:
   \[
   k - 3 = 7^a \quad \text{and} \quad k + 3 = 7^b
   \]
   where \( a \) and \( b \) are non-negative integers and \( a \leq b \). Thus:
   \[
   7^b - 7^a = 6
   \]

5. **Solving the Exponential Equation:**
   Since \( 7^a \) and \( 7^b \) are powers of \( 7 \), the only way for their difference to be \( 6 \) is if \( a = 0 \) and \( b = 1 \). Therefore:
   \[
   7^0 = 1 \quad \text{and} \quad 7^1 = 7
   \]
   This gives:
   \[
   k - 3 = 1 \quad \text{and} \quad k + 3 = 7
   \]
   Solving these, we get:
   \[
   k = 4
   \]
   Substituting \( k = 4 \) back, we get:
   \[
   7^n = 16 - 9 = 7
   \]
   Thus, \( n = 1 \).

6. **Conclusion:**
   The only solution is \( (n, p) = (1, 3) \).

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