Lemma: The sum of $k\geq 1$ consecutive positive integers is divisible by $2^j$ (ends in $j$ zeros in binary) only if either $k$ is divisible by $2^{j+1}$ or the first and last elements of the consecutive integers ia divisible by $2^{j+1}$.
Proof: Induction on $j$. For $j=0$, this means that given $k$ consecutive integers, either $k$ is even or the sum of the first and last term in the sequence is even. Basically, this means that given an odd number of consecutive binary numbers, the last binary digit of the first and last number in the sequence must be the same. I hope that is obvious.
Assume true for $j$. Now if $k$ consecutive integers add up to a multiple of $2^{j+1}$, then they add up to a multiple of $2^j$, so either $k$ is divisible by $2^{j+1}$ or the first and last element of the sequence add up to a multiple of $2^{j+1}$.
If $k$ is divisible by $2^{j+1}$, then, for any particular set of possible $j+1$ digits, there are $\frac k{2^{j+1}}$ numbers in the sequence that end in those $j+1$ digits. But we can pair off most of these equivalence classes, by pairing off the class ending in $d\pmod {2^{j+1}}$ with the class ending in $2^{j+1}-d \pmod {2^{j+1}}$ and thus, you are left with the numbers that end in $j$ or $j+1$ zeros. The ones that end in $j+1$ zeroes don't affect the last $j+1$ digits, so the sum of the numbers end in $j+1$ zeros if and only if the sum of the $\frac{k}{2^{j+1}}$ numbers that have last $j+1$ digits $10000...0$ add up to a multiple of $2^{j+1}$, which is only possible if there are an even number of them. So $\frac{k}{2^{j+1}}$ must be even, so $k$ is divisible by $2^{j+2}$
On the other hand, if the first and last number add up to a multiple of $2^{j+1}$ then $k$ must be odd, and we can pair all of the elements with it's opposite number in the sequence, except for the middle number in the sequence. But that middle number is always half of the sum of the first and last number, so for the entire sum to end in $j+1$ zeros, the average of the first and last must end in $j+1$ zeros, so the sum of the first and last must end in $j+2$ zeros.
From this lemma, you can easily prove your theorem, because $k>1$ consecutive positive integers cannot add up to $2^j$ because otherwise either the first and last add up to a multiple of $2^{j+1}$ so the largest must be bigger than $2^j$, or $k$ is divisible than $2^{j+1}$ which would make the sum far greater than $2^{j+1}$.