Notice that for any natural $n$, $\sigma(2^n)$ is odd and hence all powers of $2$ are heavy. For all other even numbers $n$ with at least 1 odd divisor (except for $1$), $\sigma(n)$ is even and hence $n$ is not heavy.
So, only powers of $2$ and odd numbers can be heavy.
The maximum consecutive string of heavy numbers is clearly present when 2 powers of $2$ have difference of exactly $2$ since otherwise, the other even number will have an odd factor.
The only powers of $2$ with difference $2$ is $2,4$.
So the string $2,3,4,5$ with length $\boxed{4}$ is maximal.
Even numbers of the form $2^x \cdot s^2$ where $s$ is odd can be heavy, but they cannot have a difference of $2$ with powers of 2, so we can dismiss them.