Answer: $5$. The quadruple $(m, n, a, k) = (1, 3, 3, 5)$ satisfies the equation. We will show the cases $k \le 4$ are not possible.

For $k = 2$, we will look at the equation in modulo $7$. $5^m \equiv a^2 \pmod{7}$ hence $m$ should be even. Therefore
$$5^m + 63n + 49 \equiv 2 \pmod{3}$$
and LHS cannot be a perfect square. Similarly the case $k = 4$ is also eliminated.

For $k = 3$, we again consider modulo $7$ and $5^m \equiv a^3 \pmod{7}$ hence $m$ should be a multiple of $3$, say $m = 3l$. Then $5^{3l} + 63n + 49 \equiv 125^l + 4 \equiv 3,5 \pmod{9}$ however $a^3 \equiv 0,1,8 \pmod{9}$ so there are no solutions in this case as well.

Since $5^m \in \{1,5,25,-1,-5,-25\} \pmod{63}$ we get
$$a^k \in \{11,24,44,48,50,54\} \pmod{63}$$
Then $a^k \equiv 2 \pmod{3}$ or $a^k \in \{3,5,6\} \pmod{7}$. Therefore, $a^k$ is not a perfect square: $k$ is odd. Moreover $a^k \in \{2,3,4,5\} \pmod{7}$ or $a^k \in \{3,5\} \pmod{9}$. Therefore, $a^k$ is not a perfect cube: $3 \nmid k$. Thus, $k \ge 5$.