Let $\pi : E \to M$ be a smooth fibre bundle with fibre $S^1$. Let $N$ be a simply connected manifold. Let $f: E \to N$ be any continuous map. Is $f$ homotopic to $g \circ \pi$ for some continuous $g : M \to N$?
The idea is that one might be able to shrink each $S^1$ fibre in the image of $f$ to a point, since $N$ is simply connected, and obtain the image of a map $g: M \to N$. Thus, the image of $f$ would potentially retract onto that of $g$ inside $N$. Would this work? I know this isn't rigorous at all, just a heuristic, but I'm struggling to come up with an actual proof for this.
As discussed in the comments, the Hopf fibration $\pi:S^3\to S^2$ gives a counterexample with the identity map $S^3\to S^3$. More generally, any $S^1$-bundle $\pi:E\to M$ where $E$ is a simply connected closed manifold will give a counterexample with the identity map $E\to E$ (which cannot factor up to homotopy through any manifold of lower dimension since $E$ is closed). To explore how you can get such examples, note that an $S^1$-bundle $E\to M$ gives an exact sequence of homotopy groups $$0\to\pi_2(E)\to \pi_2(M)\to\pi_1(S^1)\to\pi_1(E)\to\pi_1(M)\to 0$$ and so $\pi_1(E)$ will be trivial as long as $\pi_1(M)$ is trivial and the map $\pi_2(M)\to\pi_1(S^1)$ is surjective. In the case of a principal $S^1$-bundle (i.e. one whose structure group is $S^1\cong U(1)$), the map $\pi_2(M)\to\pi_1(S^1)$ can be identified with the map $\pi_2(M)\to \pi_2(BU(1))$ induced by the map $M\to BU(1)$ classifying the bundle. Moreover, since $BU(1)$ is a $K(\mathbb{Z},2)$, every homomorphism $\pi_2(M)\to\mathbb{Z}$ is induced by a map $M\to BU(1)$ if $M$ is simply connected. The upshot of all of this is that if $M$ is any simply connected closed manifold with a surjective homomorphism $\pi_2(M)\to \mathbb{Z}$, this induces a map $M\to BU(1)$ which classifies an $S^1$-bundle $\pi:E\to M$ such that $E$ is simply connected and so the identity map $E\to E$ is a counterexample to your question.