Solution:

Writing $\log_n x = \tfrac{\log x}{\log n}$ and $\log_m x = \tfrac{\log x}{\log m},$ the equation becomes a quadratic in $\log x$ whose roots sum to $\log(x_1 x_2) = \tfrac18(7\log m + 6\log n).$ Hence $N^8 = m^7 n^6,$ where $N = x_1 x_2.$ For each prime dividing $mn,$ the exponents $a, b$ must satisfy $7a + 6b \equiv 0 \pmod 8;$ minimizing the integer $N$ gives $N = 16,$ achieved uniquely at $m = 2^2 = 4$ and $n = 2^3 = 8.$ So $m + n = 12.$ Thus, the correct answer is A.