Prove that $2^n+3^n $ is never a perfect square My attempt :
If $n$ is odd, then the square must be 2  (mod 3), which is not possible.
Hence $n =2m$
$2^{2m}+3^{2m}=(2^m+a)^2$
$a^2+2^{m+1}a=3^{2m}$
$a (a+2^{m+1})=3^{2m} $
By fundamental theorem of arithmetic, 
$a=3^x $
$3^x +2^{m+1}=3^y $
$2^{m+1}=3^x (3^{y-x}-1) $
Which is not possible by Fundamental theorem of Arithmetic 
Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.

This is more convoluted than your way, but I already wrote it so I'll just leave it here :-) 
$4^m = (a-3^m)(a+3^m) \implies a-3^m = 2^p, a+3^m = 2^q$ with $p+q = 2m$. Note that $q \ge p$
Then $a = 2^p + 3^m = 2^q - 3^m  \iff 2^p (2^{q-p}-1) = 2\cdot 3^m$
But then it must be $p=1$, $q = 2m-1$, and we have 
$$2^{2(m-1)}-1 = (2^{m-1}-1)(2^{m-1}+1) = 3^m$$
Since only one of the factors in the LHS can be divisible by $3$, it means the other one must be $1$, hence we get $2^{m-1}-1 = 1 \implies m = 2$
But in this case $2^{m-1}+1 = 3 \neq 3^2 $, so this is impossible.
