The issue is the convergence of the nested radical.
Let $(x_n)$ be the sequence of iterated radicals. We'll have
$$x_n = \left\{\begin{matrix}
1 & n = 1\\
\sqrt{1+x_{n-1}} & n > 1
\end{matrix}\right.$$
A sequence can logically only converge to one unique value, i.e. $(x_n) \to x$. Take $n \to \infty$ and we have the infinite nested radical described,
$$x = \sqrt{1 + \sqrt{1+\sqrt{...}}}$$
Suppose we have some number $\alpha$ we want to express as an infinite root. A video by blackpenredpen on YouTube goes over generating such expressions, and touches on why some of these roots may seem to be equal to multiple values or have multiple solutions - namely, the implicit introduction of extraneous solutions.
We can start by saying the following: let $x = \alpha$ be a solution to our hypothetical infinite root. Then, throwing another arbitrary constant $\beta$ in, where $\beta \neq \alpha$,
$$\begin{align}
x = \alpha &\implies x - \alpha = 0 \\
&\implies (x - \alpha)(x - \beta) =0 \\
&\implies x^2 -(\beta+\alpha)x + \beta\alpha = 0 \\
&\implies x^2 = (\beta+\alpha)x - \beta \alpha \\
&\implies x = \sqrt{(\beta+\alpha)x - \beta \alpha} \\
\end{align}$$
And from here, we can use this sort of recursive definition to generate our infinitely nested radical:
$$\begin{align}
x &= \sqrt{ -\beta \alpha + ( \beta + \alpha ) x } \\
&= \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) x } } \\
&= \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) x } } } \\
&= \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) \sqrt{ -\beta \alpha + ( \beta + \alpha ) ... } } } \\
\end{align}$$
Notice how through all this, the other root $\beta$ still was introduced as an extraneous solution. And of course our chose to multiply by $(x - \beta)$ was arbitrary: we could've gone
$$\begin{align}
x = \alpha &\implies x - \alpha &= 0 \\
&\implies (x - \alpha)(x - \beta_1) &=0 \\
&\implies (x - \alpha)(x - \beta_1)(x - \beta_2) &=0 \\
&\implies (x - \alpha)(x - \beta_1)(x - \beta_2)(x - \beta_3) &=0 \\
&\implies (x - \alpha)(x - \beta_1)(x - \beta_2)(x - \beta_3)...(x - \beta_n) &=0 \\
\end{align}$$
and thus introduced $n$ arbitrary constants $\beta_1 ... \beta_n$, all not equal to $\alpha$, as these "fake solutions." (Fake in that we begin with the presumption that $\alpha$ is the only solution, so to say $\beta_k$ for any $k$ is a solution when $\beta_k \neq \alpha$ is an immediate contradiction.)
The algebra would be a pain in the butt in constructing these radicals for $n>1$, but it's doable. Our situation when presented with solving such a root, instead of constructing it, is essentially the same, but in reverse: we have no inherent idea of how many arbitrary solutions might have been generated, and indeed in a sense infinitely many were depending on your interpretation of the root and its recursion. So we always have to hearken back to what would make the root a proper solution:
That being, convergence of the iterated root to that value.
For example, for the sequence $(x_n)$ introduced initially, you can show that:
- It is monotone increasing
- It is bounded above by $\phi = (1 + \sqrt 5) / 2$
- We have $\sup(x_n) = \phi$ as well
- As a consequence of the previous statements, $(x_n) \to \phi$
Thus, $\phi$ is the only actual solution to this nested radical. Sure, you can have manipulations and such that supply other solutions - but take $x_1, x_2, x_3, x_4, ...$, take the limit as $n \to \infty$ of $(x_n)$: it's not going to approach any of them but $\phi$.