= Solution
\b[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
$$
H^j(F;\mathbb Q)=0\qquad(j>4).
$$
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 $H^*(S^4;\mathbb Q)$ 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 $a\in H^2(F;\mathbb Q)$ with
$$
d_3a=u.
$$
The class $a$ is nonzero and, by the degree bound above, has a first vanishing power $a^r=0$ for some $r\geq2$. Multiplicativity says that $d_3$ is a graded derivation. Since $|a|=2$ is even,
$$
0=d_3(a^r)=r\,u\,a^{r-1}.
$$
But $E_3^{3,2r-2}=H^3(S^3;\mathbb Q)\otimes H^{2r-2}(F;\mathbb Q)$, in which $u a^{r-1}\ne0$ 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.
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