Determine all positive integers $k$ for which there exist positive integers $n$ and $m, m\ge 2$, such that $3^k + 5^k = n^m$
1. **Case 1: \( k \) is even**

   - If \( k \) is even, then both \( 3^k \) and \( 5^k \) are perfect squares.
   - We have \( 3^k + 5^k \equiv 2 \pmod{4} \) because \( 3^k \equiv 1 \pmod{4} \) and \( 5^k \equiv 1 \pmod{4} \).
   - Since \( m \geq 2 \) and \( n \) must be even, \( n^m \equiv 0 \pmod{4} \).
   - This leads to a contradiction because \( 2 \not\equiv 0 \pmod{4} \).
   - Therefore, there are no solutions for \( k \) being even.

2. **Case 2: \( k \) is odd**

   - If \( k \) is odd, we need to consider the 2-adic valuation \( v_2 \).
   - We have \( v_2(3^k + 5^k) = 3 \) because \( 3^k \equiv 3 \pmod{8} \) and \( 5^k \equiv 5 \pmod{8} \).
   - Since \( m \geq 2 \), \( m \) must divide \( v_2(n^m) \), which implies \( m = 3 \).

3. **Subcase: \( k = 1 \)**

   - If \( k = 1 \), then \( 3^1 + 5^1 = 8 \).
   - We can write \( 8 = 2^3 \), so \( (k, n, m) = (1, 2, 3) \) is a solution.

4. **Subcase: \( k \geq 2 \)**

   - If \( k \geq 2 \), then \( 3^k \equiv 0 \pmod{9} \) because \( k \) is odd.
   - \( 5^k \) modulo 9 can be 5, 8, or 2.
   - We need \( n^3 = 3^k + 5^k \equiv 5, 8, 2 \pmod{9} \).
   - However, \( n^3 \equiv 0, 1, 8 \pmod{9} \), so \( 5^k \equiv 8 \pmod{9} \).
   - This implies \( 3 \mid k \), so \( k = 3t \) for some \( t \in \mathbb{N}^* \).

5. **Further Analysis**

   - Let \( 3^t = a \) and \( 5^t = b \), then \( 3^k + 5^k = a^3 + b^3 = n^3 \).
   - By Fermat's Last Theorem, there are no non-trivial integer solutions to \( a^3 + b^3 = n^3 \) for \( a, b, n \) positive integers.
   - Therefore, there are no solutions for \( k \geq 2 \).

Conclusion:
The only valid solution is \( (k, n, m) = (1, 2, 3) \).

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