Solution:
We have $1006^{z} > 2011^{y} > 2011$, hence $z \geq 2$. Then $1005^{x} + 2011^{y} \equiv 0 \pmod{4}$.
But $1005^{x} \equiv 1 \pmod{4}$, so $2011^{y} \equiv -1 \pmod{4} \Rightarrow y$ is odd, i.e. $2011^{y} \equiv -1 \pmod{1006}$.
Since $1005^{x} + 2011^{y} \equiv 0 \pmod{1006}$, we get $1005^{x} \equiv 1 \pmod{1006} \Rightarrow x$ is even.
Now $1005^{x} \equiv 1 \pmod{8}$ and $2011^{y} \equiv 3 \pmod{8}$, hence $1006^{z} \equiv 4 \pmod{8} \Rightarrow z = 2$.
It follows that $y < 2 \Rightarrow y = 1$ and $x = 2$. The solution is $(x, y, z) = (2, 1, 2)$.