smc

We recall that for real numbers $a$ and $b$, there are exactly $3$ ways in which we can have $a^b=1$, namely $a=1$; $b=0$ and $a\ne 0$; or $a=-1$ and $b$ is an even integer. The first case therefore gives $$\begin{array}{rl}{x}^2-x-1=1& ⟺x^2-x-2=0\\ & ⟺(x-2)(x+1)=0\\ & ⟺x=2\text{ or }x=-1.\end{array}$$ Similarly, the second case gives $x+2=0$, i.e. $x=-2$, and this indeed gives $x^2-x-1=4+2-1=5\ne 0$, so $x=-2$ is a further valid solution. Lastly, for the third case, we have $$\begin{array}{rl}{x}^2-x-1=-1& ⟺x^2-x=0\\ & ⟺x(x-1)=0\\ & ⟺x=0\text{ or }x=1,\end{array}$$ but $x=1$ would give $x+2=3$, which is odd, whereas $x=0$ gives $x+2=2$, which is even. Therefore, this case gives only one further solution, namely $x=0$. Accordingly, the possible values of $x=-2$, $-1$, $0$, or $2$, yielding a total of $\boxed{4}$ solutions.