Diophantine equation : $6^m+2^n+2=x^2$ 
Find $m,n,x\in\mathbb{N}$ such that $6^m+2^n+2=x^2$.

My first approach is to show that for $m,n\geq2$, there exist no solution for $x$ by using modulo $4$.

Case $1$ : $m=1$, $x^2=2^n+8$.
As $n\geq1\implies2\mid RHS\implies2\mid x^2\implies4\mid x^2\implies4\mid LHS\implies 4\mid 2^n\implies n\geq 2$.
The equation can be reduced into $2+2^{n-2}=\bar x^2$ where $2\bar x=x$.
If $n-2\geq2$, $LHS\equiv2$ and $RHS\equiv0,1\mod4$. Therefore $n-2<2\implies n\leq3$.
Checking for $2\leq n\leq3$, we have $m=1,n=3,x=4$ as a solution.

Case $2$ : $n=1$, $x^2=6^m+4$.
$m=1$ is not a solution, therefore $m\geq2\implies 4\mid LHS\implies2\mid x$.
The equation can be reduced into $2^{m-2}3^m+1=\bar x^2$ where $2\bar x=x$.
I do not know how to solve the problem after this step. Any hints or solution is appreciated.

$\textbf{Hint:}$Write the last equality as,
$2^{m-2}3^m=(\bar x-1)(\bar x+1)$.
Here, $gcd(\bar x+1,\bar x-1)=2$ unless $m=2$
