$12 \cdot 99$ Find all positive integers $m, n$ that satisfy $\left|12^{m}-5^{n}\right|=7$.
(Canadian Mathematical Olympiad Training Problem, 1989)
[Solution] If $5^{n}-12^{m}=7$.
Since

then
$$
\begin{array}{c}
5^{n} \equiv 1 \quad(\bmod 4) \\
12^{m} \equiv 0 \quad(\bmod 4) \\
7 \equiv 3 \quad(\bmod 4)
\end{array}
$$
$$
5^{n}-12^{m} \equiv 1 \quad(\bmod 4).
$$

Thus, $5^{n}-12^{m}=7$ cannot hold.
If
$$
12^{m}-5^{n}=7
$$

Clearly, $m=1, \quad n=1$ is a solution.
Furthermore, if $m>1$, then $n>1$ must hold, and conversely, if $n>1$, then $m>1$ must hold.
When $m>1, n>1$,
consider modulo 3.
We have
$$
\begin{aligned}
12^{m}-5^{n} & \equiv(-1)^{n+1}(\bmod 3) \\
7 & \equiv 1 \quad(\bmod 3)
\end{aligned}
$$

Thus, for $12^{m}-5^{n}=7$ to hold, we must have
$$
(-1)^{n+1} \equiv 1 \quad(\bmod 3).
$$

Therefore, $n$ must be odd, and we have
$$
12^{m}-5^{n} \equiv-5 \quad(\bmod 8),
$$

while
thus we have
$$
7 \equiv-1(\bmod 8).
$$

This is impossible.
Therefore, $12^{m}-5^{n}=7$ has no positive integer solutions $m$ and $n$ greater than 1.
Thus, the only solution to this problem is $m=1, n=1$.