Rewriting the given equation to: $$2^{3x} - 2^{2x} - 4\times 2^{x} + 4 = 0$$

Letting $y=2^x$, then the above equation becomes $$y^3 - y^2 - 4y + 4=0=> (y-1)(y+2)(y-2)=0 \implies y=1, \pm 2$$

However, $y=2^x > 0$, hence, $y=1, 2 \implies x = 0, 1$. The answer is $\boxed{2}$.