22. Let $m, n$ be integers greater than 1, and the equation in terms of $x$
$$
8 \log _{n} x \cdot \log _{m} x-7 \log _{n} x-6 \log _{m} x-2013=0
$$
has the smallest positive integer solution. Then $m+n=(\quad)$.
(A) 12
(B) 20
(C) 24
(D) 48
(E) 272
22. A.

By the change of base formula, we have
$$
8 \frac{1}{\lg n} \cdot \frac{1}{\lg m} \lg ^{2} x-\left(\frac{7}{\lg n}+\frac{6}{\lg m}\right) \lg x-2013=0 .
$$

Let $\lg x_{1}, \lg x_{2}$ be the two roots of the quadratic equation
$$
8 \frac{1}{\lg n} \cdot \frac{1}{\lg m} t^{2}-\left(\frac{7}{\lg n}+\frac{6}{\lg m}\right) t-2013=0
$$

By Vieta's formulas, we know
$$
\begin{array}{l}
\lg x_{1}+\lg x_{2}=\frac{\frac{7}{\lg n}+\frac{6}{\lg m}}{8 \cdot \frac{1}{\lg n} \cdot \frac{1}{\lg m}} \\
=\frac{7 \lg m+6 \lg n}{8}=\lg \sqrt[8]{m^{7} n^{6}} .
\end{array}
$$

Thus, $x_{1} x_{2}=\sqrt[8]{m^{7} n^{6}}$.
Since the equation has the smallest positive integer solution, and $m, n \in \mathbf{Z}$, $m>1, n>1$, to make $x_{1} x_{2}=\sqrt[8]{m^{7} n^{6}}$ the smallest positive integer, we only need to take $m=2^{2}, n=2^{3}$, at this time,
$$
m+n=4+8=12 \text {. }
$$