The answer is $k=2$. Note that if $k<2$ then for $n=2^i$ a sufficiently large integer, $(n,kn]$ does not contain a power of two, so $k\ge 2$. Now, let $k=2$. If $n=2^i$ then $kn$ is a power of 2, if not then $2^t<n<2^{t+1}$ for some integral $t$, thus $n<2^{t+1}<2n$.

I claim that the answer is $k=2$. For $k=2$, let $a=log_2(n)$. Then $n<2^{\lfloor a+1 \rfloor} \leq 2n$, hence $k=2$ works. If $k<2$, take $n=2^t$ for big $t$. Then $2^t=n$ and $kn<2n=2^{t+1}$, so it doesn't work.

If $k < 2$, then the interval $(2^n, k*2^n]$ has no power of two. Now $k = 2$ works because we take binary reps and the power of $2$ having one more digit than $n$ suffices.