More rigorous method for this elementary problem? The problem is:

Find all real values of $x$ such that $$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x=2\sqrt{3}$$

One solution I received was as follows:

$5+2\sqrt{6}$ can be expressed as $(\sqrt{3}+\sqrt{2})^2$, and $5-2\sqrt{6}$ as $(\sqrt{3}-\sqrt{2})^2$. This means we want to solve $$(\sqrt{3}+\sqrt{2})^{2x}+(\sqrt{3}-\sqrt{2})^{2x}=2\sqrt{3}$$
  for $x$. Note that $$(\sqrt{3}+\sqrt{2})+(\sqrt{3}-\sqrt{2})=2\sqrt{3}$$
  so $x=\dfrac{1}{2}$ is a solution.

The problem with the solution is that it doesn't prove that $\dfrac{1}{2}$ is the only solution, not to mention its absolute inelegancy. Is there a more mathematically rigorous way of doing this?
I don't mind if higher algebra is used. I'm currently attempting to self-teach myself group theory, so including some of that in a solution would do me good. 

$\left(\sqrt{3}+\sqrt{2}\right)^{2x}+\left(\sqrt{3}-\sqrt{2}\right)^{2x}=\left(\left(\sqrt{3}+\sqrt{2}\right)^{x}+\left(\sqrt{3}-\sqrt{2}\right)^{x}\right)^2-2$
$\therefore \left(\sqrt{3}+\sqrt{2}\right)^{x}+\left(\sqrt{3}-\sqrt{2}\right)^{x}=\sqrt{2+2\sqrt{3}}\tag{1}$
Similarly,
$\left(\sqrt{3}+\sqrt{2}\right)^{x}-\left(\sqrt{3}-\sqrt{2}\right)^{x}=\sqrt{2\sqrt{3}-2}\tag{2}$
$\begin{align}\therefore (1)+(2) \implies \left(\sqrt{3}+\sqrt{2}\right)^{x} & =\dfrac{1}{2}\left(\sqrt{2\sqrt{3}+2}+\sqrt{2\sqrt{3}-2}\right)\\&= \dfrac{1}{\sqrt{2}}\left(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}\right)\end{align}$ 
$$\color{blue}{\boxed{\begin{align}\therefore \left(\sqrt{3}+\sqrt{2}\right)^{2x} & = \dfrac{1}{{2}}\left(2\sqrt{3}+2\sqrt{2}\right)=\left(\sqrt{3}+\sqrt{2}\right)\end{align}}}$$ 
