The only value that satisfies is $n = 1175$.

First, we note that we can rewrite the recursion as
$$a_{n+1} = \frac{a_n^2 - 1 + 1}{a_n - 1} = \frac{a_n^2 - 1}{a_n - 1} + \frac{1}{a_n - 1} = a_n + 1 + \frac{1}{a_n - 1}. \quad (1)$$
Since the difference of $a_{n+1} - a_n > 1$, there is at most one $n$ that satisfies. Now we are going to show that $n = 1175$ does indeed satisfy. First, the above gives that
$$\begin{aligned} a_{1175} &= a_1 + 1174 + \frac{1}{a_1 - 1} + \frac{1}{a_2 - 1} + \dots + \frac{1}{a_{1174} - 1} \\ &> 850 + 1174 \\ &= 2024. \end{aligned}$$
More generally, it follows from (1) that $a_n \ge 849 + n$. From this we deduce that
$$\begin{aligned} \frac{1}{a_1 - 1} + \cdots + \frac{1}{a_{1174} - 1} &\le \frac{1}{849} + \frac{1}{850} + \cdots + \frac{1}{2022} \\ &< 51 \cdot \frac{1}{849} + 100 \cdot \frac{1}{900} + 200 \cdot \frac{1}{1000} + 400 \cdot \frac{1}{1200} + 400 \cdot \frac{1}{1600} + 23 \cdot \frac{1}{2000} \\ &= \frac{17}{283} + \frac{1}{9} + \frac{1}{5} + \frac{1}{3} + \frac{1}{4} + \frac{23}{2000} \\ &= \frac{17}{283} + \frac{20+36+60+45}{180} + \frac{23}{2000} \\ &< \frac{11}{180} + \frac{161}{180} + \frac{3}{180} \\ &< 1. \end{aligned}$$
Using this, we conclude that indeed
$$$$
$a_{1175} \le$ 850 + 1174 + $\frac{1}{849} + \frac{1}{850} + \cdots + \frac{1}{2022}$ < 2024 + 1. $\quad \square$