Example 26 (6th CMO Question) 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 $n \geqslant 2, z \leqslant 5 \cdot 2^{2 n}$.

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Let the positive integer tuple $(x, y, z, n)$ satisfy the equation
$$
x^{2 n+1}-y^{2 n+1}=x y z+2^{2 n+1}.
$$

It is evident that $x-y>0$ and $x$ and $y$ have the same parity.
Otherwise, if one of $x$ and $y$ is odd and the other is even, then the left side of (1) is odd, while the right side is even, leading to a contradiction.
Thus, $x-y \geqslant 2$.
(1) When $y=1, x \geqslant 4$, (1) becomes $x^{2 n+1}-1=x z+2^{2 n+1}$,
$$
x\left(x^{2 n}-z\right)=2^{2 n+1}+1.
$$

Given $n \geqslant 2, z \leqslant 5 \cdot 2^{2 n}$, and $x \geqslant 4$, we have
$$
\begin{aligned}
x\left(x^{2 n}-z\right) & \geqslant 4\left(4^{2 n}-5 \cdot 2^{2 n}\right)=2^{4 n+2}-5 \cdot 2^{2 n+2} \\
& =2^{2 n+1}\left(2^{2 n+1}-5 \cdot 2\right) \geqslant 2^{2 n+1} \cdot 22>2^{2 n+1}+1.
\end{aligned}
$$

This contradicts (2), so when $y=1, x \geqslant 4$, the equation has no solution.
(2) When $y=1, x=3$, (1) becomes $3^{2 n+1}-1=3 z+2^{2 n+1}$,
$$
z=3^{2 n}-\frac{1}{3}\left(2^{2 n}+1\right).
$$

Given $z \leqslant 5 \cdot 2^{2 n}$, we have $3^{2 n}-\frac{1}{3}\left(2^{2 n+1}+1\right) \leqslant 5 \cdot 2^{2 n}$,
$$
\begin{aligned}
3^{2 n} & \leqslant 5 \cdot 2^{2 n}+\frac{1}{3}\left(2^{2 n+1}+1\right)=2^{2 n}\left[5+\frac{1}{3}\left(2+\frac{1}{2^{2 n}}\right)\right] \\
& \leqslant 6 \cdot 2^{2 n},
\end{aligned}
$$

which implies $\left(\frac{3}{2}\right)^{2 n} \leqslant 6$. For $n \geqslant 3$, $\left(\frac{3}{2}\right)^{2 n} \geqslant\left(\frac{3}{2}\right)^{6}=\frac{729}{64}>6$, so $n \leqslant 2$. Given $n \geqslant 2$, we have $n=2$, thus
$$
z=3^{2 \times 2}-\frac{1}{3}\left(2^{2 \times 2+1}+1\right)=70.
$$

Upon verification, $(x, y, z, n)=(3,1,70,2)$ is a solution to (1).
(3) When $y \geqslant 2$, given $x-y \geqslant 2$, we have $x=4$. Also, $z \leqslant 5 \cdot 2^{2 n}, n \geqslant 2$, so
$$
\begin{aligned}
x^{2 n+1}-x y z & =x\left(x^{2 n}-y z\right) \geqslant x\left[(y+2)^{2 n}-y z\right] \\
& \geqslant x\left[y^{2 n}+4 n y^{2 n-1}+4 n(2 n-1)^{2 n-2}+\cdots+2^{2 n}-5 \cdot 2^{2 n} y\right] \\
& >x \cdot y^{2 n}+x \cdot 2^{2 n}+y\left[4 n y^{2 n-2}+4 n(2 n-1) y^{2 n-3}+\cdots-5 \cdot 2^{2 n}\right] \\
& >y^{2 n+1}+2^{2 n+1}+2^{2 n-3} y[8 n+4 n(n-1)-40] \\
& \geqslant y^{2 n+1}+2^{2 n+1},
\end{aligned}
$$

which implies $x^{2 n+1}-x y z>y^{2 n+1}+2^{2 n+1}$, contradicting (1). Therefore, when $y \geqslant 2$, the equation has no solution.
In summary, the original equation has only one positive integer solution $(x, y, z, n)=(3,1,70,2)$.