The q-analogue identity for rook placements and the Harer–Zagier formula

Let $Y(n,s)$ be the set of all Young diagrams with $n$ rows of lengths between $s$ and $n+s$. For a rook placement $C$ on a diagram $[?]$ in $Y(n,s)$, let $\operatorname{inv}(C)$ be the number of blank squares left after crossing out every square directly to the left of or above a rook, let $|\mu|$ denote the number of squares of $\mu$, and write $[m]_q$ for the $q$-integer, $[m]_q!!$ for the corresponding $q$-double factorial, and $\genfrac{[}{]}{0pt}{}{a}{b}_q$ for the $q$-binomial coefficient. The q-analogue conjecture. The following identity holds:

$$\sum_{\mu \in Y(n,s)} \prod_{i=1}^{n}q^{\mu_i-i}[\mu_i-i+1]_q = \sum_{\mu \in Y(n,s)}\sum_{\substack{\operatorname{rook placements}\\ C \,\operatorname{on}\, \mu}}q^{\operatorname{inv}(C) + |\mu| - \binom{n+1}{2}}$$ $$=\sum_{k\geq 0} q^{n(s-k) + \binom{k}{2}}[2n-1]_q!!\genfrac{[}{]}{0pt}{}{s}{k}_q\genfrac{[}{]}{0pt}{}{n}{k}_q\prod_{i=1}^k(1 + q^{n+i}).$$

This gives a $q$-analogue of the rook-placement expression for the Harer–Zagier formula; it was verified computationally up to $n=10$ and $s=5$, but no proof or general resolution is supplied in the source.