Nested radicals $\sqrt{C+1+b\sqrt{C+1+b^{2}\sqrt{C+1+b^{3}\sqrt{\cdot\cdot\cdot}}}}=g(x)$ then $\lim_{x\to\infty}(g(x+1)-g(x))=^?1$

Sketch of proof :

Let's start from :

Choose $C=0$ then define :

$$f(x)=\sqrt{x+1},h(x)=f(\sqrt{x}f(xf(x^{3/2}f(\cdots ))$$

Then we have :

$$\lim_{x\to\infty}h''(x)=0$$

For a proof See Faà di Bruno formula and induction works .

Now introducing :

$$p(x)=\frac{h(x)}{x}$$

Then :

$$p''(x)=\frac{x^{2}h''\left(x\right)-2xh'\left(x\right)+2h\left(x\right)}{x^{3}}=\frac{-2xh'\left(x\right)+2h\left(x\right)}{x^{3}}$$

Then one can show for $x>M>0$:

$$h(x)\leq x^2,\lim_{x\to \infty}h'(x)=0$$

So : $$\lim_{x\to \infty}p''(x)=0$$

And :

$$\lim_{x\to \infty}p(x)=C$$

Currently I don't know how to show that $C=1$ but there are some similarities with $$\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{...}}}}}$$

To be continued

This is not a formal proof but it will give you an idea why your conjecture should be true. Let's define a backward sequence $(a_{n})_{n\ge 0}$:
$$a_{n-1}=\sqrt{C+x^{\frac{n}{2}}a_n} \tag{1}$$ where $g(x)=a_{0}$ (I redefined the constant $C$)
Assuming $O(a_{n})=O(x^{u_{n}})$, then by $(1)$, we have:
$$O(a_{n-1})=O(x^{\frac{n}{4}})O(a_{n}^{\frac{1}{2}})\Rightarrow O(x^{u_{n-1}})=O(x^{\frac{n}{4}})O(x^{\frac{u_{n}}{2}})$$ $$\Rightarrow u_{n-1}=\frac{n}{4}+\frac{u_{n}}{2}\Rightarrow u_{0}=\frac{u_{n}}{2^{n}}+\sum_{k=1}^{n} \frac{k}{2^{k+1}}$$ Take $n\to\infty$, $\frac{u_{n}}{2^{n}}\to 0$, leaves us: $$u_{0}=\sum_{k=1}^{\infty} \frac{k}{2^{k+1}}=1\Rightarrow g(x)=a_{0}=O(x)\Rightarrow \lim_{x\to \infty} \frac{g(x)}{x}=L \in \mathbb{R}$$ There are many ways to prove that: $$\lim_{x\to \infty} (g(x+1)-g(x))=\lim_{x\to \infty} \frac{g(x)}{x}=L \tag{2}$$ From $(1)$, we see the leading coefficient before every square root is always $1$, hence $L=1$, with our previous result $(2)$, we finally get: $$\lim_{x\to \infty} (g(x+1)-g(x))=1$$ As expected.
As you see, what I did above is a quick analysis of asymptotics, a real proof will be a lot more technical and explained more carefully.