Answer: 106

Small Hint:

By change of base, each factor equals $\frac{k^2-1}{k^2-4} \cdot \frac{\log(k+1)}{\log k}$

Big Hint:

Factor $k^2 - 1 = (k-1)(k+1)$ and $k^2 - 4 = (k-2)(k+2);$ everything telescopes

Solution:

By the change-of-base formula, $\log_k (5^{k^2-1}) = \frac{(k^2 - 1)\log 5}{\log k},$ so each factor of the product equals $$\begin{gathered} \frac{\frac{k^2-1}{\log k}}{\frac{k^2-4}{\log(k+1)}} \\ = \frac{(k-1)(k+1)}{(k-2)(k+2)} \\ \quad {}\cdot \frac{\log(k+1)}{\log k}. \end{gathered}$$

All three pieces telescope over $k = 4, \ldots, 63:$ $$\prod_{k=4}^{63} \frac{k-1}{k-2} = \frac{62}{2} = 31,$$ $$\prod_{k=4}^{63} \frac{k+1}{k+2} = \frac{5}{65} = \frac{1}{13},$$ $$\begin{gathered} \prod_{k=4}^{63} \frac{\log(k+1)}{\log k} \\ = \frac{\log 64}{\log 4} = 3. \end{gathered}$$

The product is $31 \cdot \frac{1}{13} \cdot 3 = \frac{93}{13},$ which is in lowest terms, so $m + n = 93 + 13 = 106.$