Find all positive integers $m$ and $n$ such that $n^m - m$ divides $m^2 + 2m$.
To find all positive integers \( m \) and \( n \) such that \( n^m - m \) divides \( m^2 + 2m \), we will proceed step-by-step.

1. **Initial Observations:**
   - We need \( n^m - m \) to be a divisor of \( m^2 + 2m \).
   - Since \( n \) and \( m \) are positive integers, we start by considering small values of \( n \) and \( m \).

2. **Case Analysis for \( n \):**
   - **For \( n = 1 \):**
     \[
     n^m - m = 1^m - m = 1 - m
     \]
     - This is negative for \( m > 1 \) and does not divide \( m^2 + 2m \) for any \( m \geq 1 \). Hence, \( n = 1 \) is not a valid solution.

   - **For \( n = 2 \):**
     \[
     n^m - m = 2^m - m
     \]
     - We need \( 2^m - m \) to divide \( m^2 + 2m \). Let's test small values of \( m \):
       - \( m = 1 \):
         \[
         2^1 - 1 = 1 \quad \text{and} \quad 1^2 + 2 \cdot 1 = 3
         \]
         - \( 1 \) divides \( 3 \), so \( (2, 1) \) is a solution.
       - \( m = 2 \):
         \[
         2^2 - 2 = 4 - 2 = 2 \quad \text{and} \quad 2^2 + 2 \cdot 2 = 8
         \]
         - \( 2 \) divides \( 8 \), so \( (2, 2) \) is a solution.
       - \( m = 3 \):
         \[
         2^3 - 3 = 8 - 3 = 5 \quad \text{and} \quad 3^2 + 2 \cdot 3 = 15
         \]
         - \( 5 \) divides \( 15 \), so \( (2, 3) \) is a solution.
       - \( m = 4 \):
         \[
         2^4 - 4 = 16 - 4 = 12 \quad \text{and} \quad 4^2 + 2 \cdot 4 = 24
         \]
         - \( 12 \) divides \( 24 \), so \( (2, 4) \) is a solution.
       - \( m = 5 \):
         \[
         2^5 - 5 = 32 - 5 = 27 \quad \text{and} \quad 5^2 + 2 \cdot 5 = 35
         \]
         - \( 27 \) does not divide \( 35 \), so \( (2, 5) \) is not a solution.

   - **For \( n = 3 \):**
     \[
     n^m - m = 3^m - m
     \]
     - We need \( 3^m - m \) to divide \( m^2 + 2m \). Let's test small values of \( m \):
       - \( m = 1 \):
         \[
         3^1 - 1 = 2 \quad \text{and} \quad 1^2 + 2 \cdot 1 = 3
         \]
         - \( 2 \) does not divide \( 3 \), so \( (3, 1) \) is not a solution.
       - \( m = 2 \):
         \[
         3^2 - 2 = 9 - 2 = 7 \quad \text{and} \quad 2^2 + 2 \cdot 2 = 8
         \]
         - \( 7 \) does not divide \( 8 \), so \( (3, 2) \) is not a solution.
       - \( m = 3 \):
         \[
         3^3 - 3 = 27 - 3 = 24 \quad \text{and} \quad 3^2 + 2 \cdot 3 = 15
         \]
         - \( 24 \) does not divide \( 15 \), so \( (3, 3) \) is not a solution.

   - **For \( n = 4 \):**
     \[
     n^m - m = 4^m - m
     \]
     - We need \( 4^m - m \) to divide \( m^2 + 2m \). Let's test small values of \( m \):
       - \( m = 1 \):
         \[
         4^1 - 1 = 3 \quad \text{and} \quad 1^2 + 2 \cdot 1 = 3
         \]
         - \( 3 \) divides \( 3 \), so \( (4, 1) \) is a solution.
       - \( m = 2 \):
         \[
         4^2 - 2 = 16 - 2 = 14 \quad \text{and} \quad 2^2 + 2 \cdot 2 = 8
         \]
         - \( 14 \) does not divide \( 8 \), so \( (4, 2) \) is not a solution.
       - \( m = 3 \):
         \[
         4^3 - 3 = 64 - 3 = 61 \quad \text{and} \quad 3^2 + 2 \cdot 3 = 15
         \]
         - \( 61 \) does not divide \( 15 \), so \( (4, 3) \) is not a solution.

   - **For \( n \geq 5 \):**
     - For larger values of \( n \), \( n^m \) grows very quickly compared to \( m \), making it unlikely that \( n^m - m \) will divide \( m^2 + 2m \). We can verify this by testing a few values, but it becomes clear that the solutions are limited to smaller values of \( n \).

3. **Conclusion:**
   - The valid pairs \((n, m)\) that satisfy the condition are \((2, 1)\), \((2, 2)\), \((2, 3)\), \((2, 4)\), and \((4, 1)\).

The final answer is \( \boxed{ (2, 1) } \), \((2, 2)\), \((2, 3)\), \((2, 4)\), and \((4, 1)\).