Paper p.9 and Wikipedia relate the Tutte polynomial of a graph to another polynomial.

If $(x-1)(y-1)=z$, $$T(G;x,y)=F_G(z,y)/H(G)$$

Where $F_G(z,y)=\sum_{A \subseteq E}z^{c(G_A)}(y-1)^{|A|} $ where $c(G)$ is the number of connected components and $H(G)$ has simple closed form. $F_G(z,2)$ is polynomial in $z$ and the coefficient of $C$ of $z$ is the number of connected subgraphs of $G$ (not necessarily spanning). The constant coefficient of $F_G(z,2)$ is zero or one, depending on the number of connected components of the empty graph.

So $$F_G(z,2) \mod z^2= Cz+B=T(G;z+1,2) H(G) \mod z^2 \qquad (1)$$ with $B$ zero or one (known).

If $G$ is planar, $T(G;x,y)$ is efficiently computable for $z=2$ which gives the RHS of (1).

Setting $z=2$ we get $2C+B \mod 4 =\rm{known}$, which allows finding $C \mod 2$.

Will this approach work?

I can't reproduce the polynomial relation in sagemath.

Paper p.9 and Wikipedia relate the Tutte polynomial of a graph to another polynomial.

If $(x-1)(y-1)=z$, $$T(G;x,y)=F_G(z,y)/H(G)$$

Where $F_G(z,y)=\sum_{A \subseteq E}z^{c(G_A)}(y-1)^{|A|} $ where $c(G)$ is the number of connected components and $H(G)$ has simple closed form. $F_G(z,2)$ is polynomial in $z$ and the coefficient of $C$ of $z$ is the number of connected spanning subgraphs of $G$. The constant coefficient of $F_G(z,2)$ is zero.

So $$F_G(z,2) \mod z^2= Cz=T(G;z+1,2) H(G) \mod z^2 \qquad (1)$$

If $G$ is planar, $T(G;x,y)$ is efficiently computable for $z=2$ which gives the RHS of (1).

Setting $z=2$ we get: $2C \mod 4 =2 T(G,3,2)=\rm{known}$, which allows finding $C \mod 2$.

Will this approach work?

I reproduce the polynomial relation in sagemath after help on MSE.