jmc

Let $y=\sqrt[3]{1+\sqrt{x}}.$ Then $y^3=1+\sqrt{x},$ so we can write the given equation as $$\sqrt{1+\sqrt{y^3+1}}=y.$$Squaring both sides, we get $$1+\sqrt{y^3+1}=y^2,$$so $\sqrt{y^3+1}=y^2-1.$ Squaring both sides, we get $$y^3+1=y^4-2y^2+1,$$which simplifies to $y^4-y^3-2y^2=0.$ This factors as $y^2(y-2)(y+1)=0.$ Since $y=\sqrt[3]{1+\sqrt{x}}$ has to be at least one, $y=2.$ Then $$\sqrt[3]{1+\sqrt{x}}=2,$$so $1+\sqrt{x}=8.$ Then $\sqrt{x}=7,$ so $x=\boxed{49}.$