, 2023

The answer is No. Suppose by contradiction that there is such a sequence. First, we will prove by induction that $x_n > 2$ for every $n \ge 2$. One can check with $n = 2$, then assume that the assertion is true for $n = k \ge 2$, i.e. $x_k > 2$ then
$$\frac{2x_k^2 + 2}{x_k + 3} - 2 = \frac{2x_k^2 - 2x_k - 4}{x_k + 3} = \frac{2(x_k - 2)(x_k + 1)}{x_k + 3} > 0.$$
So $x_{k+1} > \frac{2x_k^2+2}{x_k+3} > 2$. Therefore, the assertion is also true for $n = k + 1$ and it is also true for all $n \ge 2$. Hence, we have
$$x_{n+1} - x_n \ge \frac{2x_n^2 + 2}{x_n + 3} - x_n = \frac{(x_n - 1)(x_n - 2)}{x_n + 3} > 0,$$
which implies that $(x_n)$ increases strictly. On the other hand, we have
$$\frac{2x_n + 2}{x_n + 3} < 2, \forall n > 0 \implies x_{n+1} < 2 + 2023 = 2025, \forall n \ge 1$$
so $(x_n)$ is upper bounded by $2025$. Hence, $(x_n)$ has a finite limit, called by $L$ and $2 < L \le 2025$. Substituting in the initial condition, we have
$$\frac{2L^2 + 2}{L + 3} \le L \Leftrightarrow 1 \le L \le 2,$$
this contradiction shows that there is no sequence $(x_n)$ satisfying the condition. $\square$