For $y>0$ define $a_n = \underbrace{\sqrt{y+\sqrt {y + \cdots+\sqrt y}}}_{n \text{ roots}}$. Clearly the sequence $(a_n)$ is increasing. Also define $b_n = \underbrace{\sqrt{y+\sqrt {y + \cdots+\sqrt{y + \ell}}}}_{n \text{ roots}}$, where $\ell = \dfrac {1 + \sqrt{1+4y}} {2}$, thus $\sqrt{y + \ell} = \ell$. Then $a_n < b_n = \ell$ for all $n$, so the sequence $(a_n)$ is bounded from above, hence convergent to a limit $L$. Since $a_{n+1}^2 = y + a_n$, passing to the limit yields $L^2 = y + L$, thus $L=\ell$. Reversing the computation, given $x>1$ one should take $y = x^2 - x$.

define $f(x)=\sqrt{y+x}: \{x\geq1\}\rightarrow\{x\geq1\}$, then apply Banach fixed point theorem. ($y>0$)