Example 20 (1992 Canadian Mathematical Olympiad Training Problem) Find the positive integer solutions $(x, y, z)$ to the equation $8^{x}+15^{y}=17^{x}$.
Since $8^{x} \equiv 0(\bmod 8), 17^{z} \equiv 1^{z} \equiv 1(\bmod 8)$, from the given equation we get $15^{y} \equiv$ $1(\bmod 8)$, which implies that $y$ is even.

Also, since $8^{x} \equiv 1(\bmod 7), 15^{y} \equiv 1(\bmod 7)$, from the original equation we get $17^{x} \equiv 2(\bmod 7)$, i.e., $3^{z} \equiv$ $2(\bmod 7)$, hence $z$ must be even.

Furthermore, since $17^{z} \equiv(-1)^{x} \equiv 1(\bmod 3), 15^{y} \equiv 0(\bmod 3)$, it follows that $8^{x} \equiv 1(\bmod 3)$, which implies that $x$ is also even.
Let $x=2 k, y=2 m, z=2 n$, then we have
$$
2^{6 k}=\left(17^{n}-15^{m}\right)\left(17^{n}+15^{m}\right),
$$

Thus, $17^{n}-15^{m}=2^{t}$,
$$
17^{n}+15^{m}=2^{6 k-t} \text {. }
$$

Since $t>0$, $2^{6 k-2 t}+1$ is odd, so from (3) we get $t=1$.
Substituting into (1) gives $17^{n}-15^{m}=2$.

If $m \geqslant 2$, then $15^{m} \equiv 0(\bmod 9)$. When $n$ is even, $17^{n} \equiv 1(\bmod 9)$, and when $n$ is odd, $17^{n} \equiv$ $8(\bmod 9)$, so (4) does not hold when $m \geqslant 2$. Therefore, (4) has a solution if and only if $m=1, n=1$. In this case, from (2) we know $k=1$. Thus, the original equation has only one unique solution $(x, y, z)=(2,2,2)$.