First, note that $(a-b)^3+(b-c)^3+(c-a)^3$ can be factored as $3(a-b)(b-c)(c-a)$. Then the equation becomes: \[\lvert (a-b)(b-c)(c-a)\rvert=2^k.\]We substitute $a-b=x$, $b-c=y$, $c-a=-x-y$. This means that $x$, $y$, and $x+y$ are all powers of 2. If $x=2^n$ and $y=2^m$, then $2^m+2^n$ is a power of 2, so $x=y$. Then: \[2^n\cdot2^n\cdot2^{n+1}=2^k\implies k=3n+1.\]