7. Let $f(m)$ be the product of the digits of the positive integer $m$. Find the positive integer solutions to the equation $f(m)=m^{2}-10 m-36$.
Let $m$ be an $n$-digit positive integer. If $n \geq 3$, then $m \geq 10^{n-1} \geq 100$, so
$$
9^{n} \geq f(m)=m^{2}-10 m-36=m(m-10)-36 \geq 90 m-36 \geq 9 \cdot 10^{n}-36>10^{n}
$$

Contradiction, thus there is no solution in this case;
(4 points)
If $n=1$, then $f(m)=m=m^{2}-10 m-36$, this equation has no integer solution;
(8 points)
If $n=2$, and $m \geq 20$, then $81 \geq f(m)=m^{2}-10 m-36 \geq 400-200-36=164$, contradiction.
Therefore, $m \in [10,19]$.
(12 points)

Let $m=10+k, f(m)=(10+k)^{2}-10(10+k)-36=k$, solving this gives $k=3$.
In summary, the solution to the equation is $m=13$.
(16 points)