Till Infinity!

The Above equation can be written as,

$\large \displaystyle \sqrt{1 + x} = x\\ \large \displaystyle \text{Squaring on Both sides}\\ \large \displaystyle \implies 1 + x = x^2\\ \large \displaystyle \implies x^2 - x - 1 = 0.\\ \large \displaystyle \text{Solving for }x, \\ \large \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{1+4}}{2}\\ \large \displaystyle \implies x = \frac{1 \pm \sqrt5}{2}\\ \large \displaystyle \text{To find the maximum value of }x,\\ \large \displaystyle x = \frac{1+\sqrt5}{2}\\ \large \displaystyle x \approx \color{#D61F06}{\boxed{1.618033988749898}}.$

Define a sequence with $a_0=1$ and $a_{n+1}=\sqrt{1+a_n}$
$x$ is the limit of this sequence.
Claim: the sequence converges to $\dfrac{1+\sqrt 5}2$
Proof:
First it is shown that if the sequence converges, then its limit is $\dfrac{1+\sqrt 5}2$ or $\dfrac{1-\sqrt 5}2$
This can be shown by observing $x=\sqrt{1+x}$ and solving it.

Second it is shown that the sequence really converges.
By induction and the recurrence definition we can show that for all $n$ , $0<a_n<2$ (it is a loose bound, but it's enough) and also, the sequence is strictly increasing.
A strictly increasing and upper bounded sequence converges.
Also, since $0<a_n$ for all $n$ , it does not converge to $\dfrac{1-\sqrt 5}2$ , but $\dfrac{1+\sqrt 5}2$ .
So $x=\dfrac{1+\sqrt 5}2$ .

The answer for this problem is famous. We called it "the golden ratio"

let $x=\sqrt{1+x}$ (if remove one square root or more from infinity it doesn't matter)

$x^2=1+x$
$x^2-x-1=0$
using the quadradic formula we can get the solution for x.
$x=\frac{-b+\sqrt{b^2-4ac}}{2a}$ (only for maximum value of x)
$x=\frac{1+\sqrt{-1^2-4\times1\times-1}}{2\times1}$
$x=\frac{1+\sqrt5}{2}$
$x=\boxed {1.618}$