[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
$$z=3^{2 n}-\frac{1}{3}\left(1+2^{2 n+1}\right) .$$

To make $z \leqslant 5 \cdot 2^{2 n}$, 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 equation (1), we get $z=70$, thus obtaining a 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
$$\begin{aligned} x^{2 n+1}-x y z & =x\left(x^{2 n}-z\right) \\ & \geqslant 4\left(\left(4^{2 n}-5 \cdot 2^{2 n}\right)\right. \\ & =2^{2 n+2}\left(2^{2 n}-5\right) \\ & >2^{2 n+1}+1 \\ & =2^{2 n+1}+y^{2 n+1} . \end{aligned}$$

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 $(3,1,70,2)$.