Loving 2 and its powers

$0=\dfrac{2^2-2^2}{2^2-2^1}, 1=\dfrac{2^2-2^1}{2^2-2^1}, 2=\dfrac{2^3-2^2}{2^2-2^1}, 3=\dfrac{2^3-2^1}{2^2-2^1}$

$4=\dfrac{2^4-2^3}{2^2-2^1}, 5=\dfrac{2^5-2^1}{2^3-2^1}, 6=\dfrac{2^4-2^2}{2^2-2^1}, 7=\dfrac{2^4-2^1}{2^2-2^1}$

$8=\dfrac{2^5-2^4}{2^2-2^1}, 9=\dfrac{2^7-2^1}{2^4-2^1}, 10=\dfrac{2^6-2^2}{2^3-2^1}$

Suppose $11$ is still possible.

$11=\dfrac{2^a-2^b}{2^c-2^d}\iff11(2^c-2^d)=2^a-2^b$

Suppose $a>b, c>d$ . Then

$11*2^d(2^{c-d}-1)=2^b(2^{a-b}-1)$

From the equation above we get: $2^d\mid2^b$ and $2^b\mid2^d$ , so $b=d$ . Furthermore

$11(2^x-1)=2^y-1$

where $x, y$ are positive integers. Since $x, y\geq2, 4\mid2^x$ and $4\mid2^y$ . From that

$11(2^x-1)\equiv1\space\mod4$

$2^y-1\equiv3\space\mod4$

This is a contradiction.

So the answer is $\boxed{11}$ .