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. 

Hint: $5+2\sqrt{6} = \dfrac{1}{5-2\sqrt{6}}$
