What is this number: $x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\approx 4.14031...$? I was watching this Mathologer video about Ramanujan's nested radical identity
$$
3 = \sqrt{1 + 2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\, ,
$$
and I decided to look at this for different sets of "radical coefficients". Using powers of $2$ yields the quantity
$$
x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\, .
$$ 
I can't figure out much about the nature of the number $x$. I can't even prove that it exists, let alone anything analytically about its value. Nevertheless, by looking at the truncation of this expression at different levels in Mathematica, it seems to be the case that:
1) The expression converges relatively quickly.
2) $x \approx 4.14031456214125981180937290...$
I tried feeding this number into an Inverse Symbolic Calculator... to no avail, although $x$ does match about 8 digits of
$$
\frac{5}{4}\, \frac{\sqrt{3} + \sqrt{5}\ln(5)}{\ln(5)}\, .
$$
Can anyone tell me anything else about this quantity?
Edited to add:
I note that the apparently simpler Nested Radical Constant
$$
\sqrt{1 + \sqrt{2 + \sqrt{3 + ...}}}
$$
has no closed form, so we clearly have no reason to expect my quantity $x$ to have a closed form either...
Edit 2:
There was an answer that someone posted, and then deleted for some reason, which gave a method for getting a lower bound on this quantity. A variation of that method leads to:
\begin{align}
x &= \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\\
&> \sqrt{1+2^1\sqrt{0+2^2\sqrt{0+2^3\sqrt{0+...}}}}\\
&= \sqrt{1+2^1 \times 2^{2/2}\times 2^{3/4}\times 2^{4/8}\times 2^{5/16}...}\\
&= \sqrt{1 + 2^{\sum_{m=0}^\infty \frac{m+1}{2^m}}}\\
&= \sqrt{1 + 2^4} = \sqrt{17} \approx 4.1231...
\end{align}
This is apparently a relatively tight lower bound.

What does
$$
x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\, .
$$
mean? I interpret it as the limit of this sequence:
\begin{align}
x_1 &= \sqrt{1 + 2^1} \\
x_2 &= \sqrt{1 + 2^1 \sqrt{1 + 2^2}} \\
x_3 &= \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3}}} \\
x_4 &= \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3 \sqrt{1 + 2^4}}}} \\
\end{align}
Comparing $x_{n+1}$ to $x_n$:
\begin{align}
x_{n+1} 
&= \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3 \dotsb \sqrt{1 + 2^n \sqrt{1 + 2^{n+1}}}}}} \\
&> \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3 \dotsb \sqrt{1 + 2^n \sqrt{1}}}}} = x_n \\
\end{align}
so the sequence is strictly increasing.
I have not found a majorant sequence yet. First numerical results seem to suggest convergence.
One can calculate $x_n$ via the sequence
\begin{align}
y_{n, 0} &= 1 \\
y_{n, k+1} &= \sqrt{1 + 2^{n-k} y_{n, k}} \quad (k \in \{ 0, \dotsc, n-1 \} ) \\
x_n &= y_{n, n}
\end{align}
This allows to interpret the evaluation of the term for $x_n$ as a series of iterations against the functions
$$
r_{n-k}(x) = \sqrt{1 + 2^{n-k} x}
$$
which varies for $d = n-k$ from $n$ down to $1$.
The images below show the iterations for $x_3$ and $x_8$:

In this view the fixed points of the $r_d$ would influence the iteration.
$$
x_1^* = 1 + \sqrt{2} = 2.4142\dotsc \\
x_2^* = 2 + \sqrt{5} = 4.2361\dotsc \\
$$
