smc

Rearranging logs, the original equation becomes $$\frac{8}{\log n\log m}(\log x{)}^2-\left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x-2013=0$$ By Vieta's Theorem, the sum of the possible values of $\log x$ is $\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n\log m}}=\frac{7\log m+6\log n}{8}=\log \sqrt[8]{m^7{n}^6}$. But the sum of the possible values of $\log x$ is the logarithm of the product of the possible values of $x$. Thus the product of the possible values of $x$ is equal to $\sqrt[8]{m^7{n}^6}$. It remains to minimize the integer value of $\sqrt[8]{m^7{n}^6}$. Since $m,n>1$, we can check that $m=2^2$ and $n=2^3$ work. Thus the answer is $4+8=\boxed{12}$.