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.

This is justa a partial solution.
we assume first that $n>8$ and we see $2^8+2^{11}+2^n=k^2$ as
$2^8(1+2^3+2^{n-8})=k^2$ we see $2^8$ is a square...
so the problem reduces to see when would $1+2^3+2^{n-8}$ be a square..
As we have assumed $n>8$ we would have $1+2^3+2^{n-8}$ to be an odd integer.
Now, any odd perfect square is congruent to $1$ modulo $8$
we see that $1+2^3$ is congruent to $1$ modulo $8$
so the problem is to make $2^{n-8}$ as $0$ modulo $8$
i.e., to make $n-8=3k$ 
least possible case would be $k=1$ i.e., $n=12$
I tried to repeat this and let $k=2$ but then It does not work.
being congruent to $1$ mod $8$ is only necessary condition but not sufficient so I can not say this works for all $k\geq 2$.
trying to solve this completely for a long time with no success. so i thought of posting what i have done and expecting for some suggestions.
