Solve $x^2 = 2^n + 3^n + 6^n$ over positive integers. 
Solve $x^2 = 2^n + 3^n + 6^n$ over positive integers.

I have found the solution $(x, n) = (7, 2)$. I have tried all $n$'s till $6$ and no other seem to be there.
Taking $\pmod{10}$, I have been able to prove that if $4|n$ that this proposition does not hold. Can you give me some hints on how to proceed with this problem?
Thanks.

If $n=2k+1\ge 3$, then $x^2\equiv 3\mod 4$.
If $n=2k\ge 4$, then $(x+2^k)(x-2^k)=x^2-2^{2k}=6^{2k}+3^{2k}=3^{2k}(1+2^{2k})$.
We have $\gcd (x-2^k, x+2^k)=1$, then 
$x+2^k\ge 3^{2k} \Rightarrow x-2^k\ge 3^{2k}-2^{2k+1}>2^{2k}+1$.
