Simply-connected rational homology spheres

In dimension 4, we have the following: Simply-connectedness implies that $H_1(M)=0$. The condition that $M$ be a rational homology sphere implies that $H_2(M), H_3(M)$ are finitely generated torsion groups. It follows that $H^3(M) = Ext(H_2(M),\mathbb{Z})$, which is noncanonically isomorphic to $H_2(M)$ again (that's true for finitely generated torsion groups).

But Poincare duality tells us that $H^3(M)=H_1(M) =0$, so $H_2(M)=0$. Similarly, we can obtain $H_3(M)=0$. It follows that $M$ is already a homology sphere.

In dimension 5, there's the first counterexample: The so-called Wu manifold $SU(3)/SO(3)$ has homology groups $\mathbb{Z}, 0, \mathbb{Z}/2, 0, 0, \mathbb{Z}$, so rationally, it is a homology sphere.

A complete answer can be found in a paper by D. Ruberman Null-homotopic embedded spheres of codimension one: a simply-connected rational homology $n$-sphere that is not homeomorphic to $S^n$ exists if and only if $n\ge 5$. See the bottom of page 230 and example 7 on p.232.

Yes, every simply-connected rational homology $4$-sphere is topologically the $4$-sphere. Simply-connected closed topological $4$-manifolds are classified by their intersection form $Q_X:H^2(X;\Bbb Z) \times H^2(X ;\Bbb Z) \to \Bbb Z$ and their Kirby-Siebenmann invariant by a famous theorem of Freedman. If the form is even, the KS invariant automatically vanishes. If $X$ is a rational homology sphere, $Q_X$ clearly vanishes (as $H^2(X;\Bbb Z)=0$), and therefore $X$ must be homeomorphic to the $4$-sphere.

See: Michael H. Freedman & Frank Quinn Topology of 4-Manifolds (PMS-39)