Consider two graphs $G_1$ and $G_2$, thought of as $1$-dimensional CW complexes. Suppose that $X = G_1\times G_2$ has the homotopy type of a manifold (with boundary). What can be said about the graphs $G_1$ and $G_2$?
I believe one of $G_k$ must be homotopic to the circle. If you consider a vertex with more than three edges, you can `pull' one of the edges along another and get a homotopic graph where that vertex has one less edge (and a new vertex with three edges). So we can reduce to the case of degree $3$ vertices. I want to argue that if neither $G_1$ nor $G_2$ is a circle, then the product will behave badly near the product of two degree $3$ vertices.
I am not entirely sure how to argue. I know that the cone on $A * B$ is the cone on $A$ times the cone on $B$, so there is a point in $X$ whose neighborhood is the cone on three points joined with three points (i.e., a cone on the utilities problem). Since $X$ is homotopy equivalent to a manifold, there is a continuous map $X\to M$ (and a map back), so this `local structure' appears around some point in a manifold, but I do not see the way to a contradiction.
Does the fact that it is a $2$-complex tell us anything via Poincare duality somehow, like the dimension of the manifold is bounded (ignoring spheres and their quotients)? Or should I be `thinking globally' somehow? Alternative solutions are acceptable, though I'd prefer some comments on how to make the above sketchy thoughts better.
Every finite CW complex is of the homotopy type of a compact manifold with boundary, by “fattening up the cells.” In particular, the product of two $S^1\vee S^1$s, viewed as graphs, is homotopy equivalent to the product of a solid two-holed torus with itself, which disproves your conjecture.