To prove boundedness, it is sufficient to show that $a_n \le 1$ for all $n \ge 3$. Rewrite the condition in the statement in the equivalent form
$$a_{n+1}^2 - 1 \le (1-a_n)(a_{n+2}-1) = (a_n-1)(1-a_{n+2}) \quad \text{for all } n \ge 1.$$
We first show that $\min(a_n, a_{n+1}) \le 1$ for all $n \ge 2$. Suppose, if possible, that $a_n > 1$ and $a_{n+1} > 1$ for some $n \ge 2$. Then
$$\begin{aligned} a_n - 1 &< a_n^2 - 1 \le (1 - a_{n-1})(a_{n+1} - 1) \le a_{n+1} - 1 < a_{n+1}^2 - 1 \le \\ &\le (a_n - 1)(1 - a_{n+2}) \le a_n - 1, \end{aligned}$$
which is a contradiction.
To reach a final contradiction, suppose $a_n > 1$ for some $n \ge 3$. By the preceding, $a_{n-1} \le 1$ and $a_{n+1} \le 1$, so $0 < a_n^2 - 1 \le (1 - a_{n-1})(a_{n+1} - 1) \le 0$, which is the desired contradiction.