Example 4 Given the sequence $\left\{a_{n}\right\}$:
$$
a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}(n \geqslant 1) \text {. }
$$

Determine the periodicity of the sequence $\left\{a_{n}\right\}$.
Solving, from $a_{1}=2$, we get
$$
a_{2}=\frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=\frac{5 a_{2}-13}{3 a_{2}-7}=1,
$$

which means $a_{1}, a_{2}, a_{3}$ are all distinct.
From the given conditions, we have
$$
\begin{array}{l}
a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}, \\
a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{7 a_{n}-13}{3 a_{n}-5}, \\
a_{n+3}=\frac{5 a_{n+2}-13}{3 a_{n+2}-7}=a_{n},
\end{array}
$$

which means $a_{n+3}=a_{n}(n=1,2, \cdots)$.
In summary, the sequence $\left\{a_{n}\right\}$ is a purely periodic sequence, with the smallest positive period being 3.