Find a positive fixed point and prove decrease above it

If $x_n\to\ell$, continuity gives $\ell=2\sqrt{\ell+1}$, with $\ell\ge0$. Squaring yields $\ell^2-4\ell-4=0$, so $\ell=2\pm2\sqrt2$; the negative root is inadmissible. Hence $\ell=2+2\sqrt2$.

Let $A=2+2\sqrt2$ and $B=2-2\sqrt2<0$. For $x>A$, $(x-A)(x-B)>0$, so $x^2>4(x+1)$. Both sides to be compared are positive, hence $x>2\sqrt{x+1}$. With $x=x_n$, this gives $x_{n+1}<x_n$. The source sign graph of $2\sqrt{x+1}-x$ shows the same negative region beyond $A$.