jmc

We first cube each side of the equation to get $x^2-4x+4=16^3$. Notice that $x^2-4x+4=(x-2{)}^2.$

Therefore, we have that $x-2=\pm 16^{3/2}=\pm 64$. Therefore, the possible values of $x$ are $-62$ and $66,$ and the only positive value is therefore $\boxed{66}$.