A new approach to find value of $x^2+\frac{1}{x^2}$

Here is a mildly disguised version of the same idea.

Let $x=e^u$

Then we have $$\cosh u=2$$ And we want $$x^2+\frac {1}{x^2}=2\cosh 2u=2(2\cosh^2 u-1)=2(8-1)=14$$

Not sure how to show what the continued fraction equals but here's my attempt. Again similar in spirit to the algebraic method.
Solving for $x$ in the first equation we have: $x = 4 - \frac{1}{x}$.
Substituting this into the second equation:
$ (4 - \frac{1}{x})^{2} + \frac{1}{(4 - \frac{1}{x})^{2}} = 16 - \frac{8}{4-\frac{1}{x}} + \frac{1}{ 16 - \frac{8}{4-\frac{1}{x}}} = 16 - \cfrac{8}{4-\cfrac{1}{4-\cfrac{1}{4-\cdots}}} + \frac{1}{16 - \cfrac{8}{4-\cfrac{1}{4-\cfrac{1}{4-\cdots}}}} $ And that big mess equals 14.