Determine all triples of positive integers $(a, m, n)$ such that $a^m +1$ divides $(a+1)^n$.
1. **Applying Zsigmondy's Theorem:**
   Zsigmondy's theorem states that for any integers \(a > 1\) and \(m > 1\), there exists a prime \(p\) that divides \(a^m + 1\) but does not divide \(a + 1\), with the exception of the cases where \((a, m) = (2, 3)\) or \(m = 2\) and \(a + 1\) is a power of 2.

2. **Case Analysis:**
   - **Case 1: \(a = 1\):**
     \[
     a^m + 1 = 1^m + 1 = 2
     \]
     \[
     (a + 1)^n = 2^n
     \]
     Since \(2\) divides \(2^n\) for any \(n \geq 1\), all triples \((1, m, n)\) are solutions.

   - **Case 2: \(a > 1\) and \(m > 1\):**
     By Zsigmondy's theorem, there exists a prime \(p\) that divides \(a^m + 1\) but does not divide \(a + 1\), except for the case \((a, m) = (2, 3)\).

3. **Special Case: \((a, m) = (2, 3)\):**
   \[
   a^m + 1 = 2^3 + 1 = 9
   \]
   \[
   (a + 1)^n = 3^n
   \]
   For \(9\) to divide \(3^n\), we need \(n \geq 2\). Thus, the triples \((2, 3, n)\) are solutions for \(n \geq 2\).

4. **Trivial Check for Other Cases:**
   For \(a > 1\) and \(m > 1\) with \((a, m) \neq (2, 3)\), Zsigmondy's theorem guarantees a prime \(p\) that divides \(a^m + 1\) but not \(a + 1\). Therefore, there are no solutions in these cases.

The final answer is \( \boxed{ (1, m, n) } \) for any \(m, n \geq 1\) and \((2, 3, n)\) for \(n \geq 2\).