We can assume $a>b$ and $c>d$ without loss of generality. Claim 1: We need $c-d|a-b$ and $b\ge d$. Proof. Note that in order for the expression to be an integer, we need $2^c-2^d=2^d(2^{c-d}-1)|2^a-2^b=2^b(2^{a-b}-1)$. By considering the factors of 2, we have $d\le b$ and by considering the other prime factors, $2^d$ and $2^b$ can't contribute prime factors that are not 2. We see, thus, that $2^{c-d}-1$ and $2^{a-b}-1$ make up all the non-2 prime factors, so we need $2^{c-d}-1|2^{a-b}-1$ which is equivalent to $c-d|a-b$. Let $x=c-d$, $y=b-d$, and $z=\frac{a-b}{c-d}$. Then, our expression is $\frac{2^x(2^{yz}-1)}{2^y-1}=2^x(1+2^z+2^{2z}+...+2^{(y-1)z})$. The way this is set up, if $n$ can be expressed iff $2n$ can be expressed, so we can just check the odd numbers. 1=1,3=1+2,5=1+4,7=1+2+4,9=1+8, but 11 has no representation. Thus, the answer is 11.