11. Proof: Squaring both sides of the equation, we get
$$\begin{aligned} z^{4}= & \left(x^{4}-4 y^{4}\right)^{2} \\ & =x^{8}-8 x^{4} y^{4}+16 y^{8} \\ & =\left(x^{8}+8 x^{4} y^{4}+16 y^{8}\right)-16 x^{4} y^{4} \\ & =\left(x^{4}+4 y^{4}\right)^{2}-(2 x y)^{4} \end{aligned}$$

Thus,
$$(2 x y)^{4}+z^{4}=\left(x^{4}+4 y^{4}\right)^{2}$$

From this, we can see: If $x_{0}, y_{0}, z_{0}$ is a positive integer solution to the equation $x^{4}-4 y^{4}=z^{2}$, then $2 x_{0} y_{0}, z_{0}, x_{0}^{4}+4 y_{0}^{4}$ is a positive integer solution to the equation $x^{4}+y^{4}=z^{2}$. From the proof of Lemma 3, we already know that $x^{4}+y^{4}=z^{2}$ has no positive integer solutions, so $x^{4}-4 y^{4}=z^{2}$ cannot have positive integer solutions.