The following proof is obtained jointly with Alexey Gordeev.

We start with a general

**Lemma.** Let $H=(X,E)$ be a $2d$-regular graph on the vertex set $X$, $G=H\square C_k$ be the product of $H$ and a cycle of length $k$. Fix a field $\mathbb{F}$ and a subset $A\subset \mathbb{F}$, $|A|=d+2$. Let $\mathcal{U}$ denote the set of all proper $d$-colorings $u:X\to A$ of the vertices of $H$ with colors taken from $A$. Consider the square matrix $M$ with rows and columns indexed by the elements of $\mathcal{U}$:
$$
M_{u,v}=f_H(u)\cdot \prod_{x\in X} \frac{u(x)-v(x)}{\prod_{b\in A\setminus u(x)} (u(x)-b)}
$$
for two proper colorings $u,v\in \mathcal{U}$. Here $f_H(u)=\prod_{(x,y)\in E} (u(y)-u(x))$ is a graph polynomial of $H$ (with somehow fixed sign.) Then
$$
\left[\prod_i x_i^{d+1}\right] f_G=\mathrm{tr}\,M^k,
$$
where the variables $x_i$ correspond to all $k\cdot |X|$ vertices of $G$.

**Proof.** Use the formula for the coefficient of $f_G$:
$$
\left[\prod_i x_i^{d+1}\right] f_G=\sum_{a_i\in A} \frac{f_G(a_1,\ldots)}{\prod_i \prod_{b\in A\setminus a_i} (a_i-b)} \quad \quad \quad \quad (1)
$$
The non-zero summands in the RHS of (1) correspond to proper colorings of
$G$. Any proper coloring of $G$ corresponds to a sequence $u_1,\ldots, u_k\in \mathcal{U}$ of the proper colorings of $H$. For such a sequence, the corresponding summand in RHS of (1) reads as
$$
M_{u_1,u_2}\cdot M_{u_2,u_3}\cdot \ldots \cdot M_{u_k,u_1}.
$$
The sum of such products is nothing but $\mathrm{tr}\, M^k$.
In our situation $H=C_{2n+1}$, $k=2m$ is even. We choose the field $\mathbb{F}=\mathbb{C}$ and the set $A=\{1,w,w^2\}$, where $w=e^{2\pi i/3}$.
The reason why $\mathrm{tr}\, M^{2m}\ne 0$ is that $M$ is a non-zero antihermitian matrix. That implies that the eigenvalues of $M$ are purely imaginary and not all of them are equal to 0, that yields $(-1)^m\mathrm{tr}\, M^{2m}> 0$.
It remains to prove that $M$ is antihermitian ($M$ is certainly non-zero). That is the point why we have chosen such a set of colors.
Let $u=[u_1,\ldots,u_{2n+1}]\in \mathcal{U}$ be a proper 3-coloring of $C_{2n+1}$ with 3 colors $1,w,w^2$. Denote by $u_i^*$ the unique element of the set $\{1,w,w^2\}\setminus \{u_i,u_{i-1}\}$.
Assuming that the sign of $f_{H}$ is chosen so that $f_H(u)=\prod_i (u_i-u_{i-1})$ (the indices are cyclic modulo $2n+1$), applying the obvious relation
$$\frac1{u_i-u_{i}^*}=\frac{u_i-u_{i-1}}
{\prod_{b\in A\setminus u_i} (u_i-b)}$$
we get
$$
M_{u,v}=\prod_{i=1}^{2n+1}\frac{u_i-v_i}{u_i-u_i^*}.\quad \quad (2)
$$
So the relation to check is
$$
M_{v,u}=-\overline{M_{u,v}}.\quad \quad \quad \quad (3)
$$
Applying (2) and substituting $\bar{z}=1/z$ for roots of unity $z=u_i,v_i,u_i^*$ we simplify (3) to the following:
$$
\prod_i \frac{u_i-u_i^*}{u_i}=-\prod_i \frac{v_i^*-v_i}{v_i^*} \quad \quad (4)
$$
provided that $u_i\ne v_i$ for all $i$ (if $u_i=v_i$ for some $i$, then $M_{u,v}=M_{v,u}=0$).
Denote $\varepsilon_i=u_i/u_{i-1}$, then $\varepsilon_i\in \{w,w^2\}$ and $u_i^*=u_i \varepsilon_i$. Analogously write $v_i^*=v_i \delta_i$ so we rewrite (4) as
$$
\prod_i(1-\varepsilon_i)=-\prod_i (1-\overline{\delta_i})=(-1)^{1+(2n+1)}\prod \delta_i^{-1}\prod(1-\delta_i).
$$
Obviously $\prod \varepsilon_i=\prod \delta_i=1$ so this in turn rewrites as
$$
\prod_i(1-\varepsilon_i)=\prod_i (1-\delta_i). \quad \quad (5)
$$
Now we have $1-\varepsilon=\pm i\sqrt{3}\varepsilon^2$ for $\varepsilon\in \{w,w^2\}$ (the signs are distinct for $w$ and $w^2$). Substituting this for $\varepsilon_i$'s and $\delta_i$'s and using $\prod \delta_i=\prod \varepsilon_i=1$, we reduce (5) to the following fact: *the total number of $\varepsilon_i$'s and $\delta_i$'s which are equal to $w$ is even.* Call the index $i$ white if $u_i=w\cdot v_i$ and black if $u_i=w^2\cdot v_i$. Then $\varepsilon_i=\delta_i$ if $i-1,i$ have the same color and $\varepsilon_i\ne \delta_i$ if $i-1,i$ have different color. Obviously there are even number of indices $i$ of the second type, thus the result.