jmc

We multiply both sides by $(x-2)(x+2),$ giving $$(x^2-3)(x-2)=2x,$$or $$x^3-2x^2-5x+6=0.$$Noting that $x=1$ is a root of this equation, we can factor the equation as $$(x-1)(x^2-x-6)=0,$$or $$(x-1)(x-3)(x+2)=0.$$The given equation is undefined for $x=-2,$ so the only solutions are $\boxed{1,3}.$