12.95 Find all positive integer solutions $(x, y, z, n)$ that satisfy the equation $x^{2 n+1}-y^{2 n+1}=x y z+2^{2 n+1}$, where $n \geqslant 2, \quad z \leqslant 5 \cdot 2^{2 n}$.
(6th China High School Mathematics Winter Camp, 1991)
[Solution] Let the positive integer tuple $(x, y, z, n)$ satisfy the equation
$$
x^{2 n+1}-y^{2 n+1}=x y z+2^{2 n+1} \text {. }
$$

Obviously, $x-y>0$, 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 the equation is odd, while the right side is even. This is impossible.
Thus,
$$
x-y \geqslant 2 \text {. }
$$
(1) When $y=1, x=3$, the equation (1) becomes
$$
\begin{array}{l}
3^{2 n+1}-1=3 z+2^{2 n+1} . \\
z=3^{2 n}-\frac{1}{3}\left(1+2^{2 n+1}\right) .
\end{array}
$$

Since $z \leqslant 5 \cdot 2^{2 n}$, we get the inequality
$$
\begin{array}{l} 
3^{2 n}-\frac{1}{3}\left(1+2^{2 n+1}\right) \leqslant 5 \cdot 2^{2 n}, \\
3^{2 n} \leqslant 5 \cdot 2^{2 n}+\frac{1}{3}\left(1+2^{2 n+1}\right) \\
\leqslant 2^{2 n}\left[5+\frac{1}{3}\left(\frac{1}{2^{2 n}}+2\right)\right] \\
\leqslant 6 \cdot 2^{2 n} .
\end{array}
$$

That is,
$$
\begin{array}{c}
\left(\frac{3}{2}\right)^{2 n} \leqslant 6, \\
n \leqslant \frac{\lg 6}{\lg 9-\lg 4}2^{2 n+1}+1,
\end{array}
$$
(2) and (3) are contradictory.

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

That is,
$$
x^{2 n+1}-x y z>y^{2 n+1}+2^{2 n+1} .
$$
(4) contradicts (1).

Therefore, when $y \geqslant 2$, the equation has no solution.
In summary, the original equation has only one unique positive integer solution:
$$
(x, y, z, n)=(3,1,70,2) .
$$