Example 5 (2002-2003 British Mathematical Olympiad) For each positive integer $n>1$, let $p(n)$ be the largest prime factor of $n$. Find all distinct positive integers $x, y, z$ that satisfy the following conditions.
(1) $x, y, z$ are in arithmetic progression;
(2) $p(x y z) \leqslant 3$.
Let's assume $x0$. Since $\left(x^{\prime}, y^{\prime}\right)=\left(z^{\prime}, y^{\prime}\right)=1$, we can set $x^{\prime}=3^{k}, z^{\prime}=3^{l}$, thus we have $3^{k}+3^{l}=2^{a+1}$. Because $k0$. Let $x^{\prime}=2^{k}, z^{\prime}=2^{l}$, then $2^{k}+2^{l}=2 \times 3^{\alpha}$, which simplifies to $2^{k-1}+2^{l-1}=3^{a}$. Given $k$ 2, we have
$$
2^{l-2}=\frac{3^{a}-1}{2}=\sum_{r=0}^{\alpha-1} 3^{r} \equiv \sum_{r=0}^{\alpha-1} 1^{r}=\alpha \equiv 0(\bmod 2),
$$

Therefore, $\alpha$ is even. Let $\alpha=2 n$, then we have
$2^{l-1}=\left(3^{n}-1\right)\left(3^{n}+1\right)$.
Thus, $3^{n}-1=2,3^{n}+1=4, n=1, l=4$.
Hence, $x^{\prime}=2, y^{\prime}=9, z^{\prime}=16$.
If $l=2$, then $\alpha=1$, so $x^{\prime}=2, y^{\prime}=3, z^{\prime}=4$.
In summary, $(x, y, z)=(h, 2 h, 3 h),(2 h, 3 h, 4 h)$ or $(2 h, 9 h, 16 h)$, where $h$ is an integer of the form $2^{a} \times 3^{b}$.