Let $\nu_p(n)$ denote the $p$-adic valuation of $n$, i.e. the highest power of $p$ dividing $n$.

Consider the following two $q$-series formed by infinite products $$\prod_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^2=\sum_{k\geq0}a_k\,q^k \qquad \text{and} \qquad \prod_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^n=\sum_{k\geq0}b_k\,q^k.$$ Both $a_k$ and $b_k$ have combinatorial interpretations in the context of partitions.

I would like to ask:

QUESTION. Is this true? If $k=j^2\geq1$ is a perfect square, then we have $\nu_2(a_k)=\nu_2(2b_k)$.