if the atlas of a topological manifold contain only one chart if I have an atlas $\mathcal{A}$ for a topological $m$-manifold $\mathcal{M}$ which consists of only one chart; a homeomorphism $\phi:U \subseteq \mathbb{R}^m \to \mathcal{M}$.
is it then true that $\mathcal{M}$ is necessarily smooth? I think so, because there are no transition functions - hence all transitions functions (vacously) are smooth

It doesn't make sense to ask whether a topological manifold is "smooth" or not. Smoothness is not a property that topological manifolds may or may not possess; it's an additional structure that must be added -- either a maximal atlas of smoothly compatible charts, or an equivalence class of such atlases, depending on your definition.
What you can say is that if $M$ is a topological manifold that has an atlas consisting of only one chart, then it does have a smooth structure -- in fact, there's a unique smooth structure for which that chart is a smooth chart. The proof is essentially the argument you sketched -- since there are no transition functions, any atlas with only one chart is automatically smoothly compatible.
This is discussed at some length in Chapter 1 of my book Introduction to Smooth Manifolds.
