If $n,k\in\mathbb N$, solve $2^8+2^{11}+2^n=k^2$. If $n,k\in\mathbb N$, solve $$2^8+2^{11}+2^n=k^2$$
It's hard for me to find an idea. Some help would be great. Thanks.

HINT:
We have $\displaystyle2^8+2^{11}=2304$
Now, $2^8+2^{11}+2^0=2304+1\ne k^2$
$\displaystyle 2^8+2^{11}+2^1=2304+2\equiv2\pmod8$, but $a^2\equiv0,1,4\pmod8$
So,  $n\ge2$ let $n=m+2$ where $m\ge0$
$\displaystyle 2^8+2^{11}+2^{m+2}=4(576+2^m)\implies 576+2^m$ must be perfect square
Like either method $m\ge2$ let $m=r+2$ where $r\ge0$
$\displaystyle576+2^{r+2}=4(144+2^r)$ 
Follow this step
One Observation :
$$(2^4)^2+2\cdot2^4\cdot2^6+(2^6)^2=(2^4+2^6)^2$$
