1. Find all positive integers $m, n$, such that $(m+n)^{m}=n^{m}+1413$
Given that positive integers $m, n$ satisfy $(m+n)^{m}=n^{m} +1413$, since $(m+n)^{m} \geqslant m^{m}+n^{m}$, it must be that $m^{m} \leqslant 1413$.

Thus, $m \leqslant 4$. At most, when the positive integer $n, n$ satisfies $(n +n)^{m}=n^{m}+1413$, $m$ cannot be an even number. Otherwise, regardless of whether $n$ is odd or even, one side of the equation would be odd while the other side would be even, leading to a contradiction. Therefore, $m$ can only be 1 or 3.
When $m=1$, for any positive integer $n$, it is impossible to have $(m+n)^{m}=n^{m}+1413$.
When $m=3$, for the equation to hold, $n$ must satisfy
$$
(3+n)^{3}=n^{3}+1413 \text {, }
$$

which simplifies to $n^{2}+3 n-154=0$.
Solving this, we get $n=11$ and $n=-14$. Since $n=-14$ does not meet the conditions, it is discarded. Therefore, the required positive integers $m, n$ are
$$
m=3, n=11 \text {. }
$$