Question: Let $M$ and $N$ be $3$-manifolds obtained by zero-surgery on (left-handed) trefoil and figure-eight knot respectively.

What is the easy way to prove that $M$ and $N$ are not homeomorphic?

Note: When they are knot homology spheres (they are both homology $S^1 \times S^2$'s), I cannot use the classical invariants.

Answer: 
As long as we're collectively throwing the kitchen sink at this, note that the Alexander polynomial of the knot is an invariant of the 0-surgered manifold. So since the figure eight and trefoil knots have different polynomials, the 0-surgeries are not homeomorphic.
This is a topological counterpart to Josh's geometric argument.

Since $H\_1(M) \cong H\_1(N) \cong \mathbb{Z}$, there is only one 6-fold cyclic cover of either of them; call them $M'$, $N'$. Both $M'$ and $N'$ are both torus bundles over the circle, and their monodromies are the sixth powers of the monodromies of $M$ and $N$. Since the monodromy of $M$ is a sixth root of the identity but the monodromy of $N$ isn't, $M'$ is $T^3$, while $N'$ is not.