If $n=1$, the relation can not hold, so assume otherwise. If $x>y$, the left hand side factors as $n^{y}\left(n^{x-y}+1\right)$ so $n^{x-y}+1$ is a power of $n$. But it leaves a remainder of 1 when divided by $n$ and is greater than 1, a contradiction. We reach a similar contradiction if $y>x$. So $y=x$ and $2 n^{x}=n^{z}$, so 2 is a power of $n$ and $n=2$. So all solutions are of the form $2^{x}+2^{x}=2^{x+1}$, which holds for all $x$. $2^{x+1}<2001$ implies $x<11$, so there are 10 solutions.