$12 \cdot 88$ Find all positive integers $m, n$, such that $(m+n)^{m}=n^{m}+1413$.
(2nd Northeast China Three Provinces Mathematics Invitational Competition, 1987)
[Solution] When positive integers $m, n$ satisfy $(m+n)^{m}=n^{m}+1413$, due to the inequality
$$
(m+n)^{m} \geqslant m^{m}+n^{m},
$$

it must be that
$$
m^{m} \leqslant 1413 \text {. }
$$

Since
$$
4^{4}=256, \quad 5^{5}=3125>1413 \text {, }
$$

it follows that
$$
m \leqslant 4 \text {. }
$$

When $m$ is even, if $n$ is odd, then $(m+n)^{m}$ is odd, and $n^{m}+1413$ is even, which is impossible.
If $n$ is even, then $(m+n)^{m}$ is even, and $n^{m}+1413$ is odd, which is also impossible.
Therefore, $m$ must be odd.
When $m=1$, for any positive integer $n$, it is impossible to have
$$
(m+n)^{m}=n+1=n^{m}+1413=n+1413 \text {. }
$$

When $m=3$, then from $(3+n)^{3}=n^{3}+1413$
we get
$$
n^{2}+3 n-154=0
$$
$n=11, \quad n=-14$ (not valid).
Thus, the required positive integers $m, n$ are
$$
m=3, \quad n=11 .
$$