Find all naturals $k$ such that $3^k+5^k$ is the power of a natural number with the exponent $\ge 2$.
1. **Assume the given equation**: Let \(3^k + 5^k = n^m\), where \(n\) and \(m\) are natural numbers and \(m \geq 2\).

2. **Determine the parity of \(n\)**: Since \(3^k\) and \(5^k\) are both odd, their sum \(3^k + 5^k\) is even. Therefore, \(n\) must be even.

3. **Check divisibility by 4**: Since \(n\) is even, \(n^m\) is divisible by 4. Thus, \(3^k + 5^k\) must be divisible by 4. Notice that:
   - \(3^k \equiv (-1)^k \pmod{4}\)
   - \(5^k \equiv 1^k \equiv 1 \pmod{4}\)

   For \(3^k + 5^k\) to be divisible by 4, \(k\) must be odd because:
   \[
   (-1)^k + 1 \equiv 0 \pmod{4} \quad \text{if and only if} \quad k \text{ is odd}
   \]

4. **Consider modulo 3**: Since \(3^k \equiv 0 \pmod{3}\), we have:
   \[
   n^m \equiv 5^k \pmod{3}
   \]
   Since \(5 \equiv 2 \pmod{3}\), we get:
   \[
   n^m \equiv 2^k \pmod{3}
   \]
   For \(n^m\) to be congruent to 2 modulo 3, \(m\) must be odd.

5. **Consider modulo 16**: For odd \(k\), we have:
   \[
   3^k \equiv 3 \pmod{16} \quad \text{and} \quad 5^k \equiv 5 \pmod{16}
   \]
   Thus:
   \[
   3^k + 5^k \equiv 3 + 5 \equiv 8 \pmod{16}
   \]
   Since \(n^m \equiv 8 \pmod{16}\) and \(n\) is even, \(n\) must be a power of 2. The only power of 2 that satisfies \(2^m \equiv 8 \pmod{16}\) is \(2^3 = 8\), so \(m = 3\).

6. **Check \(k \neq 1\)**: Suppose \(k \neq 1\). Then \(5^k \equiv n^3 \pmod{9}\). Since \(5^k \pmod{9}\) cycles through 5, 7, 8, 4, 2, 1, we need \(n^3 \equiv 5^k \pmod{9}\). The only cube residues modulo 9 are 0, 1, and 8. Thus, \(5^k \equiv 8 \pmod{9}\), which implies \(k \equiv 2 \pmod{6}\).

7. **Check modulo 7**: For \(k\) odd, \(3^k + 5^k \pmod{7}\) cycles through 0, 1, -1. Since \(k\) cannot be divisible by 3, we have a contradiction.

8. **Conclude \(k = 1\)**: The only solution is \(k = 1\), as \(3 + 5 = 8\).

The final answer is \( \boxed{ k = 1 } \).