Is it possible to utilize the convergence of the sequence $z_{n+1}=a/(1+z_n)$ to prove that the sequence $x_{n+2} = \sqrt{x_{n+1} x_n}$ is convergent?

Your version is quite long. Here is a hint for a shorter one. What about $y_n=\log{x_n}$ then $$x_{n+2}=\sqrt{x_{n+1}x_n} \Rightarrow \\ \log{x_{n+2}}=\frac{\log{x_{n+1}}+\log{x_n}}{2} \Rightarrow \\ 2y_{n+2}=y_{n+1}+y_{n}$$ which is a linear homegenious recurrence which can be solved using characteristic polynomials. I.e. characteristic polynomial is $$2x^2-x-1=0$$ with $1$ and $-\frac{1}{2}$ as the roots. Thus the general term of the sequence is $$y_n=A\cdot 1^n+B\cdot \left(-\frac{1}{2}\right)^n= A+B\cdot \left(-\frac{1}{2}\right)^n$$ or $$x_n=e^{A}\cdot e^{B\cdot \left(-\frac{1}{2}\right)^n}\to e^A, n\to\infty$$ $A$ can be found from the initial $x_0,x_1$. A few examples here, here and here.

Here is as elementary a proof as I can come up with that the limit exists and the limit is $\sqrt[3]{x_1^2 x_0} $.

$x_{n+2} = \sqrt{x_{n+1} x_n} $, so, taking logs,

$\begin{array}\\ \log x_{n+2} &= \log\sqrt{x_{n+1} x_n}\\ &= \frac12 \log(x_{n+1} x_n)\\ &= \frac12 (\log x_{n+1} +\log x_n)\\ &= \frac12 \log x_{n+1} +\frac12 \log x_n\\ \end{array} $

Letting $y_n = \log x_n$, this becomes $y_{n+2} =\frac12 (y_{n+1}+y_n) $.

$y_{n+2}-y_{n+1} =\frac12 (y_{n+1}+y_n)-y_{n+1} =-\frac12(y_{n+1}-y_n) $.

By induction, for $k > 0$,

$\begin{array}\\ y_{n+2}-y_{n+1} &=-\frac12(y_{n+1}-y_n)\\ &=\frac14(y_{n}-y_{n-1})\\ &=-\frac18(y_{n-1}-y_{n-2})\\ &...\\ &=(-1)^k\dfrac1{2^k}(y_{n+2-k}-y_{n+1-k})\\ &=(-1)^{n+1}\dfrac1{2^{n+1}}(y_{1}-y_{0}) \qquad\text{by setting }k = n+1\\ &\to 0 \qquad\text{as } n \to \infty\\ \end{array} $

This shows that $\lim_{n \to \infty} y_n$ exists.

(Actually, a bit more is needed, but the following gives an explicit value so you don't need to worry about it.)

To get the value, sum the expression for $y_{n+1}-y_n$; all the intermediate terms will cancel out.

Write it in the form $y_{n}-y_{n-1} =(-1)^{n-1}\dfrac1{2^{n-1}}(y_{1}-y_{0}) =(-1)^{n-1}\dfrac{d}{2^{n-1}} $ where $d = y_{1}-y_{0} $.

$\begin{array}\\ y_m-y_1 &=\sum_{n=2}^m (y_n-y_{n-1})\\ &=\sum_{n=2}^m (-1)^{n-1}\dfrac{d}{2^{n-1}}\\ &=d\sum_{n=2}^m (-\frac12)^{n-1}\\ &=d\sum_{n=1}^{m-1} (-\frac12)^{n}\\ &=d\dfrac{-\frac12-(-\frac12)^m}{1-(-\frac12)}\\ &=d\dfrac{-\frac12-(-\frac12)^m}{\frac32}\\ &=d\dfrac{-1-2(-\frac12)^m}{3}\\ &=d(-\dfrac13-\dfrac23(-\frac12)^m)\\ &=-\dfrac{y_1-y_0}{3}-\dfrac{2(y_1-y_0)}{3}(-\frac12)^m\\ \end{array} $

so $y_m =\dfrac{2y_1+y_0}{3}-\dfrac{2(y_1-y_0)}{3}(-\frac12)^m \to\dfrac{2y_1+y_0}{3} $.

Therefore

$\begin{array}\\ \log x_m &\to \dfrac{2y_1+y_0}{3}\\ &= \dfrac{2\log(x_1)+\log(x_0)}{3}\\ &= \dfrac{\log(x_1^2 x_0)}{3}\\ &= \log(\sqrt[3]{x_1^2 x_0})\\ \end{array} $

so $x_m \to \sqrt[3]{x_1^2 x_0} $.