Find all positive integer solutions $(x, y, z, n)$ to the following equation:
$$
x^{20+1}-y^{2 n+1}=x y z+2^{2 n+1}
$$
where $n \geqslant 2, \quad z \leqslant 5 \cdot 2^{20}$.
Solution: Clearly, $x-y>0$ and $x$ and $y$ have the same parity, let $x-y=2$.
$$
\begin{array}{l}
\text { When } y=1, x=3 \text {, let } \\
z=3^{2 n}-\frac{1}{3}\left(1+2^{2 n+1}\right) .
\end{array}
$$

If $x \leqslant 5 \cdot 2^{2 n}$, then
$$
\begin{aligned}
3^{2 n} & \leqslant 5 \cdot 2^{2 n}+\frac{1}{3}\left(1+2^{2 n+1}\right) \\
& \leqslant 2^{2 n}\left[5+\frac{1}{3}\left(\frac{1}{2^{2 n}}+2\right)\right] \\
& \leqslant 6 \cdot 2^{2 n} .
\end{aligned}
$$

From this, we get $n \leqslant 2$, and since $n \geqslant 2$, we have $n=2$. Verification shows that the equation has only one positive integer solution $(3,1,70,2)$.

When $y=1, x \geqslant 4$, since $z \leqslant 5 \cdot 2^{2 n}$ and $n \geqslant 2$, then
$$
\begin{array}{l}
x\left(x^{2 n}-z\right) \geqslant 4\left(4^{2 n}-5 \cdot 2^{2 n}\right) \\
=2^{2 n+2}\left(2^{2 n}-5\right)>2^{2 n+1}+1 .
\end{array}
$$

Thus, the equation has no solution.
When $y \geqslant 2$, since $x-y=2, z \leqslant 5 \cdot 2^{2 n}$ and $n \geqslant 2$, then
$$
\begin{array}{l}
x^{2 n+1}-x y z \geqslant x\left[(y+2)^{2 n}-y z\right] \\
=x\left[y^{2 n}+4 n y^{2 n-1}+4 n(2 n-1) y^{2 n-2}\right. \\
\left.+\cdots+2^{2 n}-5 \cdot 2^{n} y\right] \\
>x y^{2 n}+x \cdot 2^{2 n}+y\left[4 n y^{2 n-2}+4 n(2 n\right. \\
\left.-1) y^{2 n-3}-5 \cdot 2^{n}\right] \\
\quad>y^{2 n+1}+2^{2 n+1}+2^{2 n-3} y[8 n+4 n(2 n \\
\quad-1)-40] \geqslant y^{2 n+1}+2^{2 n+1} .
\end{array}
$$

Therefore, the original equation has no solution.
In summary, the original equation has a unique positive integer solution $(3,1,70,2)$.