Example 3 Given real numbers $x, y, z$ satisfy $x-y=8$, $xy+z^{2}=-16$. Then $x+y+z=$ $\qquad$ .
Let $x=a+b, y=a-b$. Then $x-y=2b \Rightarrow b=4$.
Since $xy+z^{2}=-16$, we have $(a+b)(a-b)+z^{2}+16=0$. Also, $-b^{2}=-16$, then $a^{2}+z^{2}=0$. By the property of non-negative numbers, we know $a=0, z=0$. Therefore, $x=4, y=-4$. Hence, $x+y+z=4-4+0=0$.