We rewrite the recurrence relation as follows.
$$\begin{aligned} x_{n+1} &= \sqrt{5}x_n + 2\sqrt{x_n^2 + 1} \\ \Rightarrow \quad (x_{n+1} - \sqrt{5}x_n)^2 &= 4(x_n^2 + 1) \\ \Rightarrow \quad x_n^2 - 2\sqrt{5}x_{n+1}x_n + (x_{n+1}^2 - 4) &= 0. \end{aligned}$$
This is a quadratic equation in $x_n$. We use the quadratic formula to obtain
$$x_n = \sqrt{5}x_{n+1} \pm \sqrt{5x_{n+1}^2 - (x_{n+1}^2 - 4)} = \sqrt{5}x_{n+1} \pm 2\sqrt{x_{n+1}^2 + 1}.$$
Using the given relation, it is clear that $x_{n+1} > x_n$. Therefore,
$$x_n = \sqrt{5}x_{n+1} - 2\sqrt{x_{n+1}^2 + 1}.$$
Together with $x_{n+2} = \sqrt{5}x_{n+1} + 2\sqrt{x_{n+1}^2 + 1}$, we obtain
$$x_{n+2} + x_n = 2\sqrt{5}x_{n+1}.$$
As $x_n, x_{n+1}, x_{n+2} > 0$, this relation suggests they cannot be all rational. Thus, there must be an irrational number in each of the groups $(x_{3k+1}, x_{3k+2}, x_{3k+3})$ where $k = 0, 1, \dots, 669$. That means there are at least 670 irrational numbers.