Solution:
If $n = 1$, the relation can not hold, so assume otherwise. If $x > y$, the left hand side factors as $n^{y}(n^{x-y} + 1)$ 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.