jmc

Let $y=\sqrt[4]{x}.$ Then we have $y=\frac{12}{7-y},$ or $y(7-y)=12.$ Rearranging and factoring, we get $$(y-3)(y-4)=0.$$Therefore, $y=3$ or $y=4.$ Since $x=y^4,$ we have $x=3^4=81$ or $x=4^4=256,$ so the values for $x$ are $x=\boxed{81,256}.$