jmc

Let $y=\sqrt{x^2+11},$ let $$a=\sqrt{x^2+\sqrt{x^2+11}}=\sqrt{y^2+y-11},$$and let $$b=\sqrt{x^2-\sqrt{x^2+11}}=\sqrt{y^2-y-11}.$$Then $a+b=4.$ Also, $$a^2-b^2=(y^2+y-11)-(y^2-y-11)=2y,$$and $a^2-b^2=(a+b)(a-b),$ so $$a-b=\frac{2y}{4}=\frac{y}{2}.$$Adding $a+b=4$ and $a-b=\frac{y}{2},$ we get $$2a=\frac{y}{2}+4,$$so $4a=y+8.$ Squaring both sides, we get $$16(y^2+y-11)=y^2+16y+64.$$Then $y^2=16.$ Since $y$ is positive, $y=4.$

Then $\sqrt{x^2+11}=4,$ so $x^2=5,$ and the solutions are $\boxed{\sqrt{5},-\sqrt{5}}.$ We check that these solutions work.