There are only three cases where $m^n=1$ can hold: (1) $m=1$, (2) $m\ne 0$ and $n=0$, and (3) $m=-1$ and $n$ is an even integer. - When $x^2-x-1=1$, we find $x_{1,2}=-1, 2$. - When $x+2=0$ and $x^2-x-1\ne 0$, we find $x_3=-2$. - When $x^2-x-1=-1$ and $x+2$ is even, we find $x_4=0$. Therefore the given equation has four solutions $\boxed{-2, -1, 0, 2}$.