jmc

Let $a$ and $b$ be any real numbers. Then by the Trivial Inequality, $$(a-b{)}^2\ge 0.$$This expands as $a^2-2ab+b^2\ge 0,$ so $$a^2+b^2\ge 2ab.$$(This looks like AM-GM, but we want an inequality that works with all real numbers.)

Setting $a=2xy$ and $b=x^2-y^2,$ we get $$(2xy{)}^2+(x^2-y^2{)}^2\ge 2(2xy)(x^2-y^2).$$The left-hand side simplifies to $(x^2+y^2{)}^2.$ From the given equation, $$2(2xy)(x^2-y^2)=4(xy)(x^2-y^2)=4(x^2+y^2),$$so $(x^2+y^2{)}^2\ge 4(x^2+y^2).$ Since both $x$ and $y$ are nonzero, $x^2+y^2>0,$ so we can divide both sides by $x^2+y^2$ to get $$x^2+y^2\ge 4.$$Equality occurs only when $2xy=x^2-y^2,$ or $y^2+2xy-x^2=0.$ By the quadratic formula, $$y=\frac{-2\pm \sqrt{4-4(1)(-1)}}{2}\cdot x=(-1\pm \sqrt{2})x.$$Suppose $y=(-1+\sqrt{2})x.$ Substituting into $x^2+y^2=4,$ we get $$x^2+(1-2\sqrt{2}+2)x^2=4.$$Then $(4-2\sqrt{2})x^2=4,$ so $$x^2=\frac{4}{4-2\sqrt{2}}=2+\sqrt{2}.$$So equality occurs, for instance, when $x=\sqrt{2+\sqrt{2}}$ and $y=(-1+\sqrt{2})\sqrt{2+\sqrt{2}}.$ We conclude that the minimum value is $\boxed{4}.$