What is this number: $x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\approx 4.14031...$?

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. $$ x_1^* = 1 + \sqrt{2} = 2.4142\dotsc \\ x_2^* = 2 + \sqrt{5} = 4.2361\dotsc \\ $$

Interpreting the sequence as mvw, we have $x_1<\sqrt{2^1+2^1}=2$, $x_2<\sqrt{2^1\sqrt{1+2^2}+2^1\sqrt{1+2^2}}=2\sqrt[4]{1+2^2}<2^{1+3/4}$ and similarly $$x_n<2\sqrt[4]{2^2\sqrt{\cdots\sqrt{2^n+2^n}} +2^2\sqrt{\cdots\sqrt{2^n+2^n}}}=2^{2/2+3/2^2+\cdots+(n+1)/2^n}<2^3.$$ The modus operandi is to turn each $1$ into the term that stands at the right of it, which is certainly bigger than $1$. Since $x_n$ is also increasing, its limit exists. One might as well note that the analogous sequence with powers of any fixed $m>1$ does converge, and if $m=1$ it converges to the golden ratio.

I guess my method can be refined, or there is another one, to yield a better bound.

UPDATE I thought it wouldn't be easy to majorize effectively and more sharply the sequence in a similar fashion, but I was wrong. It suffices to turn each $1$ into half the term that stands at the right of it. We thus get, for $n>1$, \begin{align} x_n &<\sqrt{(2^0+2^1)\sqrt{(2^1+2^2)\sqrt{\cdots\sqrt{2^{n-1}+2^n}}}} \\ &=2^{1/4+2/8+\cdots+(n-1)/2^n}\cdot3^{1/2+1/4+\cdots+1/2^n}<6. \end{align}

UPDATE 2 Finally nailed it! Let $P_m=\prod_{n=1}^\infty(1+2^{n+m})^{1/2^n}$. Then we find $$x<\sqrt{(1+2^1)\sqrt{(1+2^2)\sqrt{\cdots}}}=P_0<5.284,$$ $$x<\sqrt{1+2\sqrt{(1+2^2)\sqrt{(1+2^3)\sqrt{\cdots}}}}=\sqrt{1+2P_1}<4.429,$$ $$x<\sqrt{1+2\sqrt{1+2^2\sqrt{(1+2^3)\sqrt{(1+2^4)\sqrt{\cdots}}}}}=\sqrt{1+2\sqrt{1+4P_2}}<4.215 $$ etc.