The domain of $f(x)$ is $[-2, 2]$. Because $f(x)$ is continuous over this closed interval, it must reach maximum and minimal over this interval.

Let $u=\sqrt{4-x^2}$, then $u\in [0, 2]$. Accordingly,

$$f(x)=g(u)=(-u^2)^8 - 128 u = u^{16} - 128u,\qquad 0\le u \le 2$$

Setting $g'(u)$ to zero gives

$$g'(u)=16u^{15} - 128 = 0 \implies u=2^{\frac{1}{5}}\implies g\left(2^{\frac{1}{5}}\right)=-120\times 2^{\frac{1}{5}} $$

We also need to check the function values at ending points. They are

$$g(0) = 0 \qquad\text{and}\qquad g(2) = 65280$$

Therefore, the final answer is

$$\max{f} = 65280\qquad\text{and}\qquad\min{f}=-120\times 2^{\frac{1}{5}}$$