A nontrivial principal bundle which satisfies Leray-Hirsch theorem

Let $P$ be any $SU(2)$-bundle on $X$ with vanishing second Chern class $c_2(P)$. The hypotheses of the Leray-Hirsch theorem are satisfied if there is a class in $H^3(P)$ which restricts to the generator of $H^3(SU(2))$. This happens if and only if in the Leray spectral sequence, the map $d_3: H^0(X,H^3(SU(2))) \to H^4(X,H^0(SU(2))$ vanishes (since $H^i(SU(2))$ is concentrated in degrees $0$ and $3$, so the only nontrivial differentials are on the third page). This map is exactly the top Chern class.

There are nontrivial rank 2 complex bundles with vanishing Chern classes. The vanishing of $c_1$ implies that the $U(2)$ structure may be reduced to $SU(2)$, at which point the above argument shows that the hypotheses of Leray-Hirsch are satisfied.

Suppose $M = S^n$ is a sphere with $n$ odd and at least $5$. Pick your favorite Lie group $G$ for which $\pi_{n-1}(G)$ is non-trivial. (Many examples may be found here. For example, for any $n > 3$, $G= SU(\frac{1}{2}(n-1))$ works.) Since principal $G$-bundles over $M$ are classified by $[M,BG]$ which is in bijection with $[S^{n-1},G]$, there is a non-trivial principal $G$-bundle.

Suppose $P\rightarrow M$ is any such non-trivial bundle. Then $H^\ast(P;\mathbb{Q})\cong H^\ast(G;\mathbb{Q})\otimes H^\ast(M;\mathbb{Q})$. One way to see this is to note that Borel showed the universal bundle $EG\rightarrow BG$ is totally transgressive: differentials originating on the fiber are trivial, except possibly when they land in the base. Thus, the same must be true in the bundle $P\rightarrow M$. But since the rational cohomology ring of $G$ is generated in odd degrees and $H^\ast(M;\mathbb{Q})$ is concentrated in odd degrees, all the differentials must vanish.

There is also a low-dimensional example. Consider the canonical map $\mathbb RP^3 \to \mathbb RP^{\infty} \to \mathbb CP^{\infty}$, classifying the unique non-trivial principal $S^1$-bundle with base $\mathbb RP^3$. Its rational Serre spectral sequences collapses. Of course, its integral Serre spectral sequence does not collapse. The total space of this bundle is the 4-manifold $E = S^1 \times_{\mathbb Z/2} S^3$ whose fundamental group is $\mathbb Z$, not $\mathbb Z \times \mathbb Z/2\mathbb Z$.