Solution:
We will choose $x$, $y$, and $z$ to be powers of $2$ such that the terms $x^{999}$ and $y^{1000}$ are equal and $z^{1001}$ is their sum. Writing $x = 2^{a}$, $y = 2^{b}$, $z = 2^{c}$, we have the conditions
$$999a = 1000b = 1001c - 1.$$
The first equation suggests that we try $a = 1000d$, $b = 999d$, which yields the equation
$$999000d = 1001c - 1.$$
Since $999000 = 1001 \cdot 998 + 2$, we can rewrite this as
$$2d = 1001(c - 998d) - 1.$$
It is now clear that we can plug any odd positive integer $2k + 1$ for $c - 998d$ and get an integer for $d$, and hence for $c$. The results are as follows:
$$\begin{aligned} a & = 1000(1001k + 500) \\ b & = 999(1001k + 500) \\ c & = 999000k + 499001. \end{aligned}$$