Must every manifold bundle be a fiber bundle?

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I have been learning differential geometry from this lecture series by Frederic Schuller's. Here he defines a bundle (of topological manifolds) as

A triple, $(E, \pi, M),$ where $E$ and $M$ are topological manifolds and $\pi:E \longrightarrow M$ is a surjective and continuous function.

Here he defines a fiber bundle as

A bundle $(E, \pi, M)$ such that $\forall p \in M: \text{preim}_\pi \big(\{p\}\big) \cong F$ for some manifold $F$.

Using these definitions (which I hope are standard), does there exist some bundle of topological manifolds that is not a fiber bundle (i.e., has fibers that are not homeomorphic to each other)?

Sidenote, Schuller provides examples here and here of bundles over a set that are not fiber bundles, but the base set is not a manifold in either case, as he clarifies in this later video.

EDIT: I'll clarify that the $F$ in the definition of fiber bundle is intended to be the same $F$ for each point $p$, i.e., a fiber bundle is a bundle whose fibers are all homeomorphic to each other.