Example 2 Find all sequences of positive integers $\left\{a_{n}\right\}$ that satisfy the following conditions:
(1) The sequence $\left\{a_{n}\right\}$ is bounded above, i.e., there exists $M>0$, such that for any $n \in \mathbf{N}^{\cdot}$, we have $a_{n} \leqslant M ;$
(2) For any $n \in \mathbf{N}^{*}$, we have $a_{n+2}=\frac{a_{n+1}+a_{n}}{\left(a_{n}, a_{n+1}\right)}$.
Let $g_{n}=\left(a_{n}, a_{n+1}\right)$, then $g_{n} a_{n+2}=a_{n+1}+a_{n}$. Combining this with $g_{n+1}\left|a_{n+1}, g_{n+1}\right| a_{n+2}$, we know that $g_{n+1} \mid a_{n}$, which indicates that $g_{n+1}$ is a common divisor of $a_{n}$ and $a_{n+1}$, hence $g_{n+1} \leqslant g_{n}$. This means that the sequence $\left\{g_{n}\right\}_{n=1}^{+\infty}$ is a non-increasing sequence of positive integers, so there exists $N$ such that for any $n \geqslant N$, we have $g_{n}=g_{n+1}=\cdots$, denoted by $g$. Now we discuss the value of $g$.
$1^{\circ}$ If $g=1$, then for any $n \geqslant N$, we have $a_{n+2}>a_{n+1}$, i.e., $a_{n+2} \geqslant a_{n+1}+1$, thus for any $m \in \mathbf{N}^{*}$, we have
$$a_{N+m} \geqslant a_{N+1}+m-1$$

When $m$ is sufficiently large, it will contradict condition (1).
$2^{\circ}$ If $g>2$, i.e., $g \geqslant 3$, then for any $n \geqslant N$, we have $a_{n+2}<\max \left\{a_{n+1}, a_{n}\right\}$, thus for any $m \in \mathbf{N}^{*}$, we have
$$a_{N+m+2} \leqslant \max \left\{a_{N+1}, a_{N}\right\}-m$$
When $m$ is sufficiently large, the sequence $\left\{a_{n}\right\}$ will contain negative integers, which is also a contradiction.
Therefore, $g=2$. In this case, when $n \geqslant N$, we have $a_{n}=a_{n+1}$ (if not, suppose there exists $n \geqslant N$ such that $a_{n} \neq a_{n+1}$, then $\min \left\{a_{n}, a_{n+1}\right\}<a_{n+2}<\max \left\{a_{n}, a_{n+1}\right\}$, from which we can deduce that for any $m \in \mathbf{N}^{*}$, we have $\min \left\{a_{n}, a_{n+1}\right\}<a_{n+m+1}<\max \left\{a_{n}, a_{n+1}\right\}$, which is impossible), further, combining $g=2$, we know that for any $n \geqslant N$, we have $a_{n}=2$, from $a_{N+1}=a_{N}=2, g_{N}=2$ we can deduce backwards that $a_{N-1}=2, g_{N-1}=2$, and so on, we can deduce that for any $n \in \mathbf{N}^{*}$, we have $a_{n}=2$.

In summary, the sequence $\left\{a_{n}\right\}$ that satisfies the conditions is unique, and its general term is $a_{n}=2$.