 Solution

False. Take $M=S^4$ and $N=S^3$. Both are closed oriented simply connected manifolds and $4>3$. A cohomological obstruction rules out even a topological fiber bundle.

Suppose a fiber bundle $F\to S^4\to S^3$ existed. The long exact sequence of homotopy groups of a fibration, including its component terms, makes $F$ path-connected. A local trivialization over a small open ball in $S^3$ identifies its preimage, an open subset of $S^4$, with $F\times B^3$. Projection onto $F\times \{0\}$ gives a retraction. An open smooth four-manifold has the homotopy type of a CW complex of dimension at most four, so its rational cohomology vanishes in degrees above four. A retract's cohomology injects into that of the ambient space. Therefore

Use the multiplicative cohomological Serre spectral sequence with rational coefficients. Since $S^3$ is simply connected, there is no nontrivial coefficient action, and only columns zero and three occur on $E_2$. Thus $E_3=E_2$, and the only possibly nonzero differential is $d_3$. Let $u$ be the degree-three base generator. It cannot survive, because the sequence converges to $H^{\ast }(S^4;\mathbb{Q})$ and that ring has zero degree-three part. It has no outgoing differential and its only possible incoming differential is from $E_3^{0,2}$. Hence there is $a\in H^2(F;\mathbb{Q})$ withThe class $a$ is nonzero and, by the degree bound above, has a first vanishing power $a^r=0$ for some $r\ge 2$. Multiplicativity says that $d_3$ is a graded derivation. Since $|a|=2$ is even,But $E_3^{3,2r-2}=H^3(S^3;\mathbb{Q})\otimes H^{2r-2}(F;\mathbb{Q})$, in which $u{a}^{r-1}\ne 0$ by minimality of $r$; also $r$ is nonzero over $\mathbb{Q}$. This is a contradiction. Thus no such fiber bundle exists. This is the special case of a sphere cannot fibre over a lower-dimensional odd sphere required here.