Perfect Square!

$2^{11} + 2^{8} + 2^{n} = m^2$

If $n≥8$
$2^{8}(8 + 1 + 2^{n-8}) = m^2$

$16\sqrt{9 + 2^{n-8}} = m$

Then $9 + 2^{n-8} = x^2$

$2^{n-8} = (x+3)(x-3)$

$x + 3 = 2^{a}$ and $x - 3 = 2^{b}$

$2^{a} - 2^{b} = 6$

The unique pairs (a, b) are $a = 3$ and $b = 1$

Then $x=5 \rightarrow 2^{n-8} = 5^2 - 9$

$2^{n-8} = 16 \rightarrow n - 8 = 4 \rightarrow n=12$

Finally $2^{11} + 2^{8} + 2^{12} = (80)^2$

I want a binomio square: I think as $\ (2^{11}+2^{8}+2^n)=( 2*2^{10}+(2^{4})^2+(2^{6})^2)$

is the only soluction