Let $m, n, a, k$ be positive integers and $k>1$ such that the equality $$5^m+63n+49=a^k$$
holds. Find the minimum value of $k$.
To find the minimum value of \( k \) such that \( 5^m + 63n + 49 = a^k \) holds for positive integers \( m, n, a, k \) with \( k > 1 \), we will analyze the cases for \( k = 2, 3, 4 \) and show that they do not have solutions. Then, we will verify that \( k = 5 \) is a valid solution.

1. **Case \( k = 2 \):**
   \[
   5^m + 63n + 49 = a^2
   \]
   Consider the equation modulo 3:
   \[
   5^m + 63n + 49 \equiv (-1)^m + 0 + 1 \equiv (-1)^m + 1 \pmod{3}
   \]
   Since \( a^2 \equiv 0, 1 \pmod{3} \), we need \( (-1)^m + 1 \equiv 0 \text{ or } 1 \pmod{3} \). This implies:
   \[
   (-1)^m + 1 \equiv 0 \pmod{3} \implies (-1)^m \equiv -1 \pmod{3} \implies m \text{ is odd}
   \]
   Now consider the equation modulo 7:
   \[
   5^m + 63n + 49 \equiv 5^m + 0 + 0 \equiv 5^m \pmod{7}
   \]
   The quadratic residues modulo 7 are \( 0, 1, 2, 4 \). The powers of 5 modulo 7 are:
   \[
   5^1 \equiv 5, \quad 5^2 \equiv 4, \quad 5^3 \equiv 6, \quad 5^4 \equiv 2, \quad 5^5 \equiv 3, \quad 5^6 \equiv 1 \pmod{7}
   \]
   Since \( 5^m \equiv 5, 4, 6, 2, 3, 1 \pmod{7} \) and none of these are quadratic residues modulo 7, there is no solution for \( k = 2 \).

2. **Case \( k = 3 \):**
   \[
   5^m + 63n + 49 = a^3
   \]
   Consider the equation modulo 7:
   \[
   5^m + 63n + 49 \equiv 5^m \pmod{7}
   \]
   The cubic residues modulo 7 are \( 0, 1, 6 \). The powers of 5 modulo 7 are:
   \[
   5^1 \equiv 5, \quad 5^2 \equiv 4, \quad 5^3 \equiv 6, \quad 5^4 \equiv 2, \quad 5^5 \equiv 3, \quad 5^6 \equiv 1 \pmod{7}
   \]
   Since \( 5^m \equiv 5, 4, 6, 2, 3, 1 \pmod{7} \) and the only cubic residue among these is 6, we must have \( 5^m \equiv 6 \pmod{7} \). This implies \( m \equiv 3 \pmod{6} \).

   Now consider the equation modulo 9:
   \[
   5^m + 63n + 49 \equiv (-1)^r + 4 \equiv a^3 \pmod{9}
   \]
   The cubic residues modulo 9 are \( 0, 1, 8 \). Since \( (-1)^r + 4 \equiv 3 \text{ or } 5 \pmod{9} \), and neither 3 nor 5 are cubic residues modulo 9, there is no solution for \( k = 3 \).

3. **Case \( k = 4 \):**
   Since \( k = 2 \) does not have a solution, \( k = 4 \) also does not have a solution because \( a^4 = (a^2)^2 \) and we already established that \( a^2 \) cannot satisfy the equation.

4. **Case \( k = 5 \):**
   We need to verify if \( k = 5 \) is a valid solution. Consider the example given:
   \[
   m = 1, \quad n = 3, \quad a = 3, \quad k = 5
   \]
   Substituting these values into the equation:
   \[
   5^1 + 63 \cdot 3 + 49 = 5 + 189 + 49 = 243 = 3^5
   \]
   This satisfies the equation, so \( k = 5 \) is a valid solution.

The final answer is \( \boxed{5} \)