Notice that any power of $1$ is $1$, so $x^2-x-1=1$ would give valid solutions. 
$x^2-x-2=0$
$(x-2)(x+1)=0$
$x=2, -1$
Also, $-1$ to an even power also gives $1$, so we check $x^2-x-1=-1$
$x^2-x=0$
$x(x-1)=0$
$x=0, 1$
However, $x=1$ gives an odd power of $-1$, so this is discarded. Finally, notice that anything to the $0\text{th}$ power (except for $0$, as $0^0$ is undefined) gives $1$.
$x+2=0$
$x=-2$
This doesn't make $x^2-x-1=0$, so this is also valid.
Overall, our valid solutions are $x=-2, -1, 0, 2$ for a grand total of $4, \boxed{\text{C}}$.
How many integers $x$ satisfy the equation $(x^2-x-1)^{x+2}=1?$
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of these}$