(4) Find all positive integer solutions $(x, y, z, n)$ to the equation $x^{2 n+1}-y^{2 n+1}=x y z+2^{2 n+1}$, where $n \geqslant 2$ and $y \leqslant 5 \cdot 2^{2 n}$.
First, it is easy to see that $x, y$ must have the same parity, and since the right side is positive, we know $x>y$, thus $x-y \geqslant 2$.

Let $y=1$, then $x \geqslant 3$, since $(2 x) \geqslant 6$ and $z \leqslant 5 \cdot 2^{2 n}$, if there exists a positive integer solution $(x, 1, z, n)$ that satisfies the given equation, then we should have
$$
\begin{aligned}
1= & x^{2 n+1}-2^{2 n+1}-x z= \\
& (x-2)\left(x^{2 n}+x^{2 n-1} \cdot 2+\cdots+x^{n} \cdot 2^{n}+\cdots+\right. \\
& \left.x \cdot 2^{2 n-1}+2^{2 n}\right)-x z \geqslant \\
& (2 n+1) \cdot(2 x)^{n}-x z \geqslant \\
& (2 n+1) \cdot 6^{n-1} \cdot 2 x-x \cdot 5 \cdot 2^{2 n}= \\
& 2^{2 n} x\left(\left(\frac{3}{2}\right)^{n-1}\left(n+\frac{1}{2}\right)-5\right)
\end{aligned}
$$

However, for $n \geqslant 2$, for this inequality to hold, it must be that $\left(\frac{3}{2}\right)^{n-1}\left(n+\frac{1}{2}\right)3, x y \geqslant 15$, which is impossible. Therefore, when $y \geqslant 3$, the given equation has no solution.

Hence, the only positive integer solution to the given equation is $(x, y, z, n)=$ $(3,1,70,2)$.