Example 1 (2005 Russian Mathematical Olympiad) Find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers.
The smallest positive integer sought is 11.
We have:
$1=\frac{4-2}{4-2}, \quad 3=\frac{8-2}{4-2}, \quad 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2}$,
$7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2}$,
$2=2 \cdot 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \cdot 5=\frac{2^{6}-2^{2}}{2^{3}-2}$.
Assume
$11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$,
without loss of generality, we can assume $a>b, c>d$, let $m=a-b, n=c-d, k=b-d$, then we get $11\left(2^{n}-1\right)=2^{k}\left(2^{m}-1\right)$.
The left side of the above equation is odd, so $k=0$. It is easy to see that $n=1$ cannot satisfy the equation. If $m>n>1$, then $2^{n}-1$ and $2^{m}-1$ both have a remainder of 3 when divided by 4, thus the left side of the equation has a remainder of 1 when divided by 4, while the right side has a remainder of 3, which is a contradiction.