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} $.