We have $1006^{z}>2011^{y}>2011$, hence $z \geq 2$. Then $1005^{x}+2011^{y} \equiv 0(\bmod 4)$.

But $1005^{x} \equiv 1(\bmod 4)$, so $2011^{y} \equiv-1(\bmod 4) \Rightarrow y$ is odd, i.e. $2011^{y} \equiv-1(\bmod 1006)$.

Since $1005^{x}+2011^{y} \equiv 0(\bmod 1006)$, we get $1005^{x} \equiv 1(\bmod 1006) \Rightarrow x$ is even.

Now $1005^{x} \equiv 1(\bmod 8)$ and $2011^{y} \equiv 3(\bmod 8)$, hence $1006^{z} \equiv 4(\bmod 8) \Rightarrow z=2$.

It follows that $y<2 \Rightarrow y=1$ and $x=2$. The solution is $(x, y, z)=(2,1,2)$.