Answer     $4$

We study the function $$f(x)=\sqrt{(x+4) + x\sqrt{(x+5)+(x+1)\sqrt{(x+6)+\cdots}}}$$

If we can simplify $f(x)$, the answer to the original question will be $f(2)$.

This function can be re-written as $$f(x)=\sqrt{x+4 + xf(x+1)} \implies f^2(x)=x+4 + xf(x+1)$$

Let the degree of $f(x)$ be $k$. Then the degree of the left side is $2k$ and the right side is $k+1$, implying $k=1$. It follows that $f(x) = ax + b$ where $a$ and $b$ are to be determined coefficiencies. $$(ax+b)^2 = x+4 + x(a(x+1)+b)\implies \begin{cases}2ab&=&1+a+b\\b^2&=&4\end{cases}\implies (a,b)=(1, 2), \left(\frac{1}{5}, -2\right)$$

Meanwhile, we know $f(0)=2$, hence the solution must be $$(a, b)=(1,2)\implies f(x)=x+2\implies f(2)=4$$