Example 7, find all positive integer solutions $(x, y, z)$ of the equation
$$
(3 x+1)(3 y+1)(3 z+1)=34 x y z
$$

(2010, German Mathematical Olympiad)
Let's assume $x \geqslant y \geqslant z$.
If $z>\frac{1}{\sqrt[3]{34}-3}$, then $3 z+1<\sqrt[3]{34} z$.
Similarly, $3 x+1<\sqrt[3]{34} x, 3 y+1<\sqrt[3]{34} y$.
Multiplying the three inequalities, we get
$$
(3 x+1)(3 y+1)(3 z+1)<34 x y z \text {. }
$$

This contradicts the given equation.
Therefore, $z \leqslant \frac{1}{\sqrt[3]{34}-3}<5$.
(1) $z=1$.

Then $4(3 x+1)(3 y+1)=34 x y$
$$
\Rightarrow x y+6(x+y)+2=0 \text {. }
$$

Clearly, there are no positive integer solutions.
$$
\text { (2) } z=2 \text {. }
$$

Then $7(3 x+1)(3 y+1)=68 x y$
$$
\begin{array}{l}
\Rightarrow 5 x y-21(x+y)-7=0 \\
\Rightarrow(5 x-21)(5 y-21)=476=2^{2} \times 7 \times 17 \\
\Rightarrow(x, y)=(28,5) \text { or }(11,7) .
\end{array}
$$
(3) $z=3$.

Then $10(3 x+1)(3 y+1)=102 x y$
$$
\Rightarrow 6 x y-15(x+y)-5=0 \text {. }
$$

Clearly, there are no positive integer solutions.
$$
\text { (4) } z=4 \text {. }
$$

Then $13(3 x+1)(3 y+1)=136 x y$
$$
\begin{array}{l}
\Rightarrow 19 x y-39(x+y)-13=0 \\
\Rightarrow(19 x-39)(19 y-39)=1768 .
\end{array}
$$

By enumeration, there are no solutions.
In summary, $(x, y, z)=(28,5,2),(11,7,2)$.
Due to the symmetry of $x, y, z$, different permutations yield 12 solutions.