Find $x\in \Bbb R,$ solving $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$

We know that the Golden ratio $\phi = \frac{1+\sqrt5}{2}$ and $~\bar{\phi}= \frac{1 -\sqrt5}{2}$ satisfy the equation $$x^2 = 1+x \Longleftrightarrow x= \sqrt {1 + x}~~x>0~ \implies x= f(x)$$ where, $f(x)= \sqrt {1 + x},~~x>0.$ This means that $ \phi $ is the only fix points of the function $f(x)= \sqrt {1 + x},~x>0$

But $$ \sqrt {1 + \sqrt {1 + \sqrt{1+x}}} = f\circ f\circ f(x)= f^3(x) .$$

Therefore your equation reduces to $$x= f^3(x)$$

Which mean that $x$ is fix point of $f^3$.

On the other hand, $|f'(x)| =\frac{1}{2\sqrt{x+1}} \le \frac12$ then,

$$|(f^3(x))'| = 3|f'(x)\cdot f'(f^2)(x)| \le \frac 34<1$$

then $f^3$ is a contraction and hence, $f^3$ has a unique fix point satisfying $x=f^3(x)$ . Whereas, $$ \color{red}{x= f^3(x) \implies f(x) = f^3(f(x)) \implies x= f(x) }$$

since we observse that $f(x)$ is also a point fix of $f^3$ by unicity? we get $x=f(x).$

But $\phi $ is the only positive number satisfying $x=f(x)$.

Hence, the only solution to $x=f^3(x)$ is $$x=\phi = \frac{1+\sqrt5}{2}~$$

Clearly any solution of the given equation is a positive real number.

Lemma. The map $x\mapsto \sqrt{1+x}$ is a contraction over $[0,+\infty)$, since for every $x\geq 0$ we have $\frac{d}{dx}\sqrt{x+1}\leq\frac{1}{2}<1$.

Corollary. By the Banach fixed point Theorem, $x=\sqrt{1+x}$ has a unique real solution. Simple algebra gives that such a solution is provided by the golden ratio $\varphi=\frac{1+\sqrt{5}}{2}$.

Lemma. Since $f(x)$ is a contraction over $[0,+\infty)$, $f(f(f(x)))$ is a contraction a fortiori.

Corollary. By the Banach fixed point Theorem, the given equation has a unique real solution. $x=\varphi$ is a solution, hence it is the only one.

Essentially the solution is unique(which you can find out yourself). Then there is another function has the same solution to this one:

$x=\sqrt{1+x} $

Done