When is $2^n +3^n + 6^n$ a perfect square? Find all $n$ for which $2^n+3^n+6^n$ is a perfect square.
 I  do not have a specific idea how to solve this one

If $n$ is odd then $2^n+3^n+6^n\equiv -1\pmod{3}$, so $2^n+3^n+6^n$ cannot be a square.
For similar reasons $\!\!\pmod{5}$, $n$ has to be a number of the form $4k+2$.
For similar reasons $\!\!\pmod{7}$, $n$ has to be a number of the form $12k\pm 2$, so $2^n+3^n+6^n$ is a multiple of $7$. $n=2$ gives the solution:
$$ 2^2+3^2+6^2 = 7^2. $$
Let we set $a_n=2^n+3^n+6^n$ and $\nu_7(m)=\max\{n\in\mathbb{N}:7^n\mid m\}$. 
Since $\frac{1}{4}+\frac{1}{9}+\frac{1}{36}=\frac{7}{18}$, if $n=12k-2$ then $\nu_7(a_n)\equiv 1\pmod{2}$, so $a_n$ cannot be a square and it is enough to study the case $n=12k+2$ and the sequence given by $b_n=a_{12n+2}$.
My bet now is on some variation of Zsigmondy's theorem. Numerical evidence gives that every $b_n$, for $n>0$, has a big prime divisor that appears with multiplicity one. If we manage to prove the last claim, we have that the previous solution ($n=2$) is the only one.
