Is there a way to solve TSP in polynomial time if there are only two kinds of weights, 0 and 1?
No, since if every edge has weight 1, there is still the question of whether any such tour exists, which is the Hamiltonian Cycle problem, and this is still NP-hard. (The link is to a Wikipedia page for Hamiltonian Path -- both the path and cycle versions of the problem are hard.)
The accepted answer isn't quite right. An instance of TSP consists of a distance between every pair of cities: that is, it consists of a weighted complete graph. Every complete graph has a Hamiltonian cycle.
However, it is simple to reduce HAMILTON-CYCLE to $0$–$1$ TSP. Given a graph $G$, create a TSP instance where the cities are the vertices and the distance is $0$ if there is an edge between the cities and $1$ if there is not. Then $G$ has a Hamiltonian cyle if, and only if, the TSP instance has a tour of weight zero. Therefore, $0$–$1$ TSP is NP-complete.