First Round

Let $n>2$ be an integer. We want $n^{n-2}$ to be an $n$th power of some integer, i.e., there exists an integer $k$ such that $n^{n-2}=k^n$.

This means that $n^{n-2}$ is a perfect $n$th power. Let us write $n^{n-2}=(n^a{)}^n$ for some integer $a$.

But $(n^a{)}^n=n^{an}$, so we need $an=n-2$, i.e., $a=\frac{n-2}{n}$.

But $a$ is an integer, so $n$ divides $n-2$. Since $n>2$, $n$ divides $n-2$ implies $n$ divides $-2$.

The positive divisors of $2$ are $1$ and $2$, but $n>2$, so the only possibility is $n=2$ (which is excluded), or $n=-1$ or $n=-2$ (which are not $>2$).

Alternatively, $n$ divides $-2$, so $n=1$ or $2$ (excluded), or $n=-1$, $-2$ (excluded).

Therefore, there are no integers $n>2$ such that $n^{n-2}$ is an $n$th power of an integer.

Answer: There are no such integers $n>2$.