Find the smallest natural number $ k$ for which there exists natural numbers $ m$ and $ n$ such that $ 1324 \plus{} 279m \plus{} 5^n$ is $ k$-th power of some natural number.
1. We need to find the smallest natural number \( k \) such that there exist natural numbers \( m \) and \( n \) for which \( 1324 + 279m + 5^n \) is a \( k \)-th power of some natural number.
2. First, we consider \( k = 2 \) (i.e., we are looking for a perfect square).

   - Taking modulo 9:
     \[
     1324 \equiv 1 \pmod{9}, \quad 279m \equiv 0 \pmod{9}, \quad 5^n \equiv 5, 7, 8, 4, 2, 1 \pmod{9} \text{ for } n = 1, 2, 3, 4, 5, 6
     \]
     The quadratic residues modulo 9 are \( 0, 1, 4, 7 \). Therefore, we need \( 1 + 5^n \equiv 0, 1, 4, 7 \pmod{9} \).

     - For \( 5^n \equiv 8 \pmod{9} \), we get \( 1 + 8 \equiv 0 \pmod{9} \), so \( n = 3 + 6k = 3(1 + 2k) \).

   - Taking modulo 31:
     \[
     1324 \equiv 22 \pmod{31}, \quad 279m \equiv 0 \pmod{31}, \quad 5^n \equiv 5, 25, 1 \pmod{31} \text{ for } n = 1, 2, 3
     \]
     We need \( 22 + 5^n \) to be a quadratic residue modulo 31. The quadratic residues modulo 31 are \( 0, 1, 4, 5, 6, 9, 10, 13, 16, 17, 18, 20, 21, 24, 25, 28 \).

     - For \( 5^n \equiv 5 \pmod{31} \), we get \( 22 + 5 \equiv 27 \pmod{31} \), but 27 is not a quadratic residue modulo 31.
     - For \( 5^n \equiv 25 \pmod{31} \), we get \( 22 + 25 \equiv 16 \pmod{31} \), and 16 is a quadratic residue modulo 31.
     - For \( 5^n \equiv 1 \pmod{31} \), we get \( 22 + 1 \equiv 23 \pmod{31} \), but 23 is not a quadratic residue modulo 31.

     Since we have a contradiction, \( k = 2 \) is not possible.

3. Next, we consider \( k = 3 \) (i.e., we are looking for a perfect cube).

   - We need to find \( m \) and \( n \) such that \( 1324 + 279m + 5^n \) is a perfect cube.
   - Let's try \( m = 1 \) and \( n = 3 \):
     \[
     1324 + 279 \cdot 1 + 5^3 = 1324 + 279 + 125 = 1728
     \]
     \[
     1728 = 12^3
     \]
     Therefore, \( 1324 + 279 \cdot 1 + 5^3 = 12^3 \), which is a perfect cube.

Thus, the smallest \( k \) for which there exist natural numbers \( m \) and \( n \) such that \( 1324 + 279m + 5^n \) is a \( k \)-th power of some natural number is \( k = 3 \).

The final answer is \( \boxed{3} \)