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: 
They can also be distinguished geometrically. Both knots are genus one fibered knots, so both $M$ and $N$ are torus bundles over the circle.

The complement of the figure eight is hyperbolic, so the monodromy of the fibration is pseudo-Anosov. Hence the monodromy on $N$ is Anosov, and $N$ has Sol geometry.

The complement of the trefoil is Seifert fibered with base orbifold a disk with cone points of order 2 and 3. The longitude intersects an ordinary circle fiber on the boundary 6 times, so $M$ is Seifert fibered over the sphere with cone points of order 2, 3, and 6.

Since $M$ and $N$ are closed manifolds admitting distinct geometric structures, they are not homeomorphic.

Furthermore, it is possible to check that the monodromy of $M$ as a torus bundle has order 6, which is finite, so $M$ in fact admits a Euclidean structure.
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.