Will you do trial and error method?

$2^{11}+2^{8}+2^{n}=2^{8}(2^{3}+1+2^{n-8})=2^{8}(9+2^{n-8})$ .

Since $2^{8}$ is a perfect square, $9+2^{n-8}$ must also be a perfect square.

This is equivalent to determining a Pythagorean triple where one term is $3$ and where the other term is a power of $2$ .

We find that the only Pythagorean triplet satisfying this condition is $(3,4,5)$ .

Therefore, $2^{n-8}=16=2^{4} \implies n=12$ is the only solution.

Call this square $x²,~~x\in\mathbb{Z}$ . So:

$2^{11}+2^8+2^n=x²$ $2⁸(2³+1)+2^n=x²$ $2^n=x²-(2⁴.3)^2=(x+48)(x-48)$

Giving $2^m=x-48$ :

$2^m=x-48$ $2^{n-m}=x+48$

$\Longrightarrow 2^{n-m}-2^m=96$

The unique solution is $2⁷-2⁵$ . Hence:

$m=5\Longrightarrow n-m=7\iff\boxed{n=12}$