As noted in the comment by user44191, one needs only check this for prime powers. Note that this is trivial for $q=2$ and so we may assume that we have an odd prime $q$. Then the claim is that when $k \equiv \ell$ one has that $$v_2(\sigma_k(q^m)) = v_2\sigma_\ell (q^m).$$

This is the same as asserting that $$v_2(1+q^{k} + q^{2k} \cdots q^{mk}) = v_2(1+q^{\ell} + q^{2\ell} \cdots q^{m\ell}) $$

or equivalently that $$v_2\left( \frac{(q^k)^m-1}{q^k-1}\right) = v_2\left( \frac{(q^\ell)^m-1}{q^\ell-1}\right). $$

And that's the same as $$v_2\left( \frac{(q^m)^k-1}{q^k-1}\right) = v_2\left( \frac{(q^m)^\ell-1}{q^\ell-1}\right). $$

But this is true not just for prime $q$ but in fact for all odd $q$ because $\phi(2^n)$ is always a power of 2, and if $x \equiv 1$ (mod $2^n$) then $x^2 \equiv 1$ (mod $2^{n+1}$).