Proof: Let $(x, y)=m$, and set $x=m n, y=m l$, where $m, l, n$ are all positive integers, and $(l, n)=1$. Thus, from the condition we have
$$(l+n) z=m l n .$$

Using $(x, y, z)=1$, we know $(m, z)=1$, hence, from (1) we know $z \mid ln$. Since $(l, n)=1$, it follows that $(l, l+n)=1, (n, l+n)=1$, therefore, from (1) we know $l|z, n| z$, and by $(l, n)=1$, we have $ln \mid z$. So, $z=ln$, and thus $m=l+n$. This way, we have
$$\begin{array}{c} x+y=m(l+n)=(l+n)^{2} \\ x-z=m n-l n=n(m-l)=n^{2} \\ y-z=m l-l n=l(m-n)=l^{2} \end{array}$$

The proposition is proved.