jmc

Let $y=\sqrt{(2+\sqrt{3}{)}^x}.$ Then $$\sqrt{(2-\sqrt{3}{)}^x}=\sqrt{{\left(\frac{1}{2+\sqrt{3}}\right)}^x}=\frac{1}{\sqrt{(2+\sqrt{3}{)}^x}}=\frac{1}{y},$$so the given equation becomes $y+\frac{1}{y}=4.$ Then $y^2+1=4y,$ or $$y^2-4y+1=0.$$By the quadratic formula, $$y=2\pm \sqrt{3}.$$Thus, $$\sqrt{(2+\sqrt{3}{)}^x}=2\pm \sqrt{3}.$$For the $+$ root, $$\sqrt{(2+\sqrt{3}{)}^x}=2+\sqrt{3},$$so $x=2.$ For the $-$ root, $$\sqrt{(2+\sqrt{3}{)}^x}=2-\sqrt{3}=\frac{1}{2+\sqrt{3}}=(2+\sqrt{3}{)}^{-1},$$so $x=-2.$ Thus, the solutions are $\boxed{2,-2}.$