$4 \cdot 172$ Find all positive integer solutions $(x, y, z, n)$ of the equation
$$
x^{2 n+1}-y^{2 n+1}=x y z+2^{2 n+1}
$$

satisfying the conditions $n \geqslant 2, z \leqslant 5 \cdot 2^{2 n}$.
[Solution] From the given equation, it is easy to see that $x-y>0$ and $x$ and $y$ have the same parity, so $x-y \geqslant 2$.
When $y=1, x=3$, from the equation we get
$$
z=3^{2 n}-\frac{1}{3}\left(1+2^{2 n+1}\right) \text {. }
$$

To ensure $z \leqslant 5 \cdot 2^{2 n}$, we should have
$$
\begin{aligned}
3^{2 n} & \leqslant 5 \cdot 2^{2 n}+\frac{1}{3}\left(1+2^{2 n+1}\right) \\
& =\left(5+\frac{2}{3}\right) 2^{2 n}+\frac{1}{3} \leqslant 6 \cdot 2^{2 n} .
\end{aligned}
$$

This yields $n \leqslant 2$. Since the problem requires $n \geqslant 2$, we have $n=2$.
Substituting $n=2$ into (1), we get $z=70$, thus obtaining one solution to the original equation: $(3,1,70,2)$.
Next, we will prove that the equation has only this solution that meets the requirements.

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

Thus, the equation has no solution in this case.
When $y \geqslant 2$, since $x-y \geqslant 2, z \leqslant 5 \cdot 2^{2 n}, n \geqslant 2$.

Therefore,
$$
\begin{aligned}
& x^{2 n+1}-x y z \geqslant x\left[(y+2)^{2 n}-y z\right] \\
= & x\left[y^{2 n}+4 n y^{2 n-1}+4 n(2 n-1) y^{2 n-2}+\cdots+2^{2 n}-y z\right] \\
> & x y^{2 n}+x \cdot 2^{2 n}+y\left[4 n y^{2 n-2}+4 n(2 n-1) y^{2 n-3}-5 \cdot 2^{2 n}\right] \\
> & y^{2 n+1}+2^{2 n+1}+2^{2 n-3} y[8 n+4 n(2 n-1)-40] \\
\geqslant & y^{2 n+1}+2^{2 n+1} .
\end{aligned}
$$

This means that the equation has no solution when $y \geqslant 2$.
In summary, the only solution to the original equation that meets the requirements is $(3,1,70,2)$.