Example 4 Find all positive integers $a, b$ such that
$$(a, b)+9[a, b]+9(a+b)=7 a b .$$
Let $(a, b)=d$, set $a=d x, b=d y$, then $(x, y)=1$ (by property 11), $[a, b]=d x y$ (by property 8), thus substituting into (1) we get
$$\begin{array}{c}
1+9 x y+9(x+y)=7 d x y \\
7 d=9+9\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{x y}
\end{array}$$

Therefore,
$$9<7 d \leqslant 9+9\left(\frac{1}{1}+\frac{1}{1}\right)+\frac{1}{1 \times 1}=28$$

Hence,
$$2 \leqslant d \leqslant 4$$

When $d=2$, from (2) we get
$$5 x y-9(x+y)=1$$

Multiplying both sides by 5 and factoring the left side, we get
$$(5 x-9)(5 y-9)=86=2 \times 43$$

Thus, $(5 x-9,5 y-9)=(1,86), (86,1), (2,43), (43,2)$. Solving each case, we find that only $(x, y)=(2,19), (19,2)$, corresponding to $(a, b)=(4,38), (38,4)$.

Similarly discussing for $d=3,4$, we get $(a, b)=(4,4)$.
Therefore, the pairs $(a, b)$ that satisfy the condition are $(4,38), (38,4), (4,4)$.