Find all positive integer pairs $(m, n)$ such that $m- n$ is a positive prime number and $mn$ is a perfect square. Justify your answer.
To find all positive integer pairs \((m, n)\) such that \(m - n\) is a positive prime number and \(mn\) is a perfect square, we will analyze the given conditions step by step.

1. **Condition Analysis**:
   - \(m - n\) is a positive prime number, say \(p\).
   - \(mn\) is a perfect square.

2. **Case 1: \(m - n = 2\)**:
   - If \(m - n = 2\), then \(m = n + 2\).
   - For \(mn\) to be a perfect square, both \(m\) and \(n\) must share the same prime factors in such a way that their product is a perfect square.
   - If both \(m\) and \(n\) are odd, their difference \(2\) cannot be a prime number since the difference of two odd numbers is even and greater than 2.
   - If both \(m\) and \(n\) are even, then \(m\) and \(n\) must be divisible by 2. However, for \(mn\) to be a perfect square, they must share more factors, which is not possible with a difference of 2.

3. **Case 2: \(m - n = p\) where \(p\) is an odd prime**:
   - Let \(m = n + p\).
   - For \(mn\) to be a perfect square, \(m\) and \(n\) must be such that their product is a perfect square.
   - Consider \(m = x^2\) and \(n = y^2\) where \(x\) and \(y\) are integers. Then \(x^2 - y^2 = p\).
   - Factoring the difference of squares, we get \((x - y)(x + y) = p\).

4. **Prime Factorization**:
   - Since \(p\) is a prime number, the only factor pairs are \((1, p)\) and \((p, 1)\).
   - Therefore, \(x - y = 1\) and \(x + y = p\).

5. **Solving the System of Equations**:
   - From \(x - y = 1\) and \(x + y = p\), we can solve for \(x\) and \(y\):
     \[
     x - y = 1 \quad \text{(1)}
     \]
     \[
     x + y = p \quad \text{(2)}
     \]
   - Adding equations (1) and (2):
     \[
     2x = p + 1 \implies x = \frac{p + 1}{2}
     \]
   - Subtracting equation (1) from (2):
     \[
     2y = p - 1 \implies y = \frac{p - 1}{2}
     \]

6. **Verification**:
   - For \(x\) and \(y\) to be integers, \(p + 1\) and \(p - 1\) must both be even, which is true since \(p\) is an odd prime.
   - Therefore, \(x = \frac{p + 1}{2}\) and \(y = \frac{p - 1}{2}\) are integers.

7. **Constructing \(m\) and \(n\)**:
   - \(m = x^2 = \left(\frac{p + 1}{2}\right)^2\)
   - \(n = y^2 = \left(\frac{p - 1}{2}\right)^2\)

8. **Conclusion**:
   - The pairs \((m, n)\) are given by:
     \[
     m = \left(\frac{p + 1}{2}\right)^2, \quad n = \left(\frac{p - 1}{2}\right)^2
     \]
   - For each odd prime \(p\), this gives a valid pair \((m, n)\).

The final answer is \( \boxed{ (m, n) = \left(\left(\frac{p + 1}{2}\right)^2, \left(\frac{p - 1}{2}\right)^2\right) } \) for each odd prime \(p\).