Example 7 Find all pairs of positive integers $(a, b)$ such that
$$a^{3}+6 a b+1, b^{3}+6 a b+1,$$

are both perfect cubes.
Let's assume $a \leqslant b$, then
$$b^{3}<b^{3}+6 a b+1 \leqslant b^{3}+6 b^{2}+1<(b+2)^{3}$$

Since $b^{3}+6 a b+1$ is a perfect cube, we have
$$b^{3}+6 a b+1=(b+1)^{3},$$

which implies $\square$
$$\begin{aligned}
6 a b & =3 b^{2}+3 b \\
b & =2 a-1
\end{aligned}$$

Thus, $\square$
$$a^{3}+6 a b+1=a^{3}+12 a^{2}-6 a+1$$

Notice that $\quad(a+1)^{3} \leqslant a^{3}+12 a^{2}-6 a+1<(a+4)^{3}$, therefore, since $a^{3}+12 a^{2}-6 a+1$ is a perfect cube, it can only be
$$a^{3}+12 a^{2}-6 a+1=(a+1)^{3},(a+2)^{3},(a+3)^{3}$$

Solving each case, we find that the only solution is $a=1$.
Therefore, the pair $(a, b)=(1,1)$ satisfies the condition.