Perfect square......with additions?

Let $x^2=2^8+2^{11}+2^n$ , where $x$ is an integer. Thus, $x=\sqrt{2^8+2^{11}+2^n}$ $=\sqrt{2^8\left(1+2^3+2^{n-8}\right)}$ $=\sqrt{2^8}\cdot\sqrt{1+2^3+2^{n-8}}$ Since $\sqrt{2^8}$ is an integer, $\sqrt{1+2^3+2^{n-8}}$ will be also an integer as well. $1+2^3+2^{n-8}$ $=9+2^{n-8}$ Then, try a few perfect squares bigger than 9 and substitute in this equation to find $x$ . $\color{#3D99F6}\text{If it equals to 16:}$ $9+2^{n-8}=16$ $2^{n-8}=7$ $\color{#3D99F6}\text{7 is not a power of 2. Thus, this cannot be equal to 16.}$ $\color{#20A900}\text{If it equals to 25:}$ $9+2^{n-8}=25$ $2^{n-8}=16$ $2^{n-8}=2^4$ $n-8=4$ $\color{#20A900}\text{n=12}$

$2^8+2^{11}=48^2$ ,let the perfect square $=a^2$ ,then $a^2-48^2=(a+48)(a-48)=2^n$ .
Let $a+48=2^b$ and $a-48=2^c$ , $2^b-2^c=96,2^{b-5}-2^{c-5}=3$ .
The only solution is $2^2-2^0=3,b=7,c=5,b+c=n=\boxed{12}$

Note that 2^8 + 2^11 = 256 + 2048 = 2304 = 48^2. So the problem becomes to find the smallest n such that 48^2 + 2^n = x^2, or 2^n = (x - 48)(x + 48). Let x - 48 = 2^t, and x + 48 = 2^(n - t). Therefore 96 = (2^t) (2^(n - 2t) - 1) = 32*3, So t = 5, and 2^(n - 10) - 1 = 3, or 2^(n - 10) = 4 = 2^2, so n - 10 = 2, or n = 12. Ed Gray