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.
 A: Assume that $m$ and $n$ are both greater than $1$. Then, we have:
$$x^2 \equiv 6^m+2^n+2 \equiv 0+0+2 \equiv 2 \pmod{4}$$
which is impossible. Thus, we either have $m=1$ or $n=1$.

Case $1$ : $m=1$
Substituting yields:
$$2^n+8=x^2$$
If $n>3$, then $8 \mid x^2$ but $16 \nmid x^2$ which would be a contradiction. Thus, $n \leqslant 3$. It is clear by plugging in values $n \leqslant 3$, that the only solution is:
$$(m,n,x)=(1,3,4)$$

Case $2$ : $n=1$
Substituting yields:
$$6^m+4=x^2 \implies6^m=x^2-4=(x-2)(x+2)$$
Thus, the values $x-2$ and $x+2$ must contain only powers of $2$ and $3$. Clearly, only one of the factors is divisible by $3$, and is thus divisible by $3^m$. The other factor is atleast $3^m-4$, which yields:
$$6^m \geqslant 3^m(3^m-4) \implies 2^m \geqslant 3^m-4 \implies m \leqslant 2$$
Plugging in both $m=1$ and $m=2$ shows that no such solutions exist.
Thus, the only solution is $(m,n,x)=(1,3,4)$.
A: $\textbf{Hint:}$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 $m=2$
