If $M$ is topological space (lets say connected Hausdorff and second countable, because I don't think it matters particularly for the question) then we can give $M$ a smooth structure by specifying an open cover $\mathcal U = \{U_i : i \in I \}$ and homeomorphisms $\psi_i\colon U_i \to V_i \subseteq \mathbb R^n$ from the open sets in the cover to open sets $V_i$ in $\mathbb R^n$. (If one cares, we can assume that the $V_i$ all equal $ \{x \in \mathbb R^n: (\sum_{i=1}^n x_i^2)^{1/2} <1 \}$). This data must then satisfy the conditions that, for all $i,j \in I$, the maps $$ \psi_j\circ\psi_i^{-1}\colon \psi_i(U_i\cap U_j)\to \psi_j(U_i\cap U_j) \tag{$\dagger$} $$ are smooth. This last makes sense because $\psi_i(U_i\cap U_j)$ and $\psi_j(U_i\cap U_j)$ are open subsets of $\mathbb R^n$.
This is a "smooth structure" because, for example, it allows us to define what it means for a function $f\colon N\to \mathbb R$ to be smooth, where $f$ is a function on an open subset $N$ of $M$: $f$ is smooth if $f_{|N\cap U_i}$ is smooth for all $i$, and $f_{|N\cap U_i}$ is smooth if $f\circ (\psi_i)^{-1}_{|\psi(N)}\colon \psi(N)\to \mathbb R$ is smooth. (Condition $(\dagger)$ and the chain rule ensure this is a consistent definition.)
Now many texts then introduce the notion of a maximal atlas, that being an atlas $\mathcal A = (U_i,\psi_i)_{i \in I}$ such that if $\phi\colon W\to \mathbb R^n$ is a homeomorphism onto its image from an open subset $W$ of $M$ which satisfies the compatibility condition that $\theta\circ \psi^{-1}$ is a diffeomorphism for all $\psi_i \in \mathcal A$ then $(w,\theta) \in\mathcal A$. One then declares the "differentiable structure" to be this maximal atlas.
My issue is that this seems a rather pointless definition. It is, in essence, saying that the differentiable structure is the set of charts which are compatible with the atlas that you start with. That is fine, but I would argue, useless: If I want to specify the structure, then I can say "the maximal atlas containing this atlas $\mathcal A = (U_i,\psi_i)_{i \in I}$", but then to check anything about the smooth structure, I have I have to consult back to $\mathcal A$ anyway. Moreover, for manifolds like $\mathbb R^n$, it is basically just garbage -- if I want to check if $f\colon \mathbb R^n\to X$ is smooth, I don't need any chart other than the identity map.
To try and be a bit more scientific about my object, it doesn't seem a particularly natural choice of definition from the categorical point of view: From that (Yoneda) perspective, one might be inclined to say that the "differentiable structure" is the set of all smooth maps $f\colon M\to N$ as $N$ runs over all smooth manifolds. This is, obviously, also a huge amount of data (though it is not like a maximal atlas is small gadget), but since the notion of a smooth structure is local, it is enough to understand the smooth maps from $M$ to $\mathbb R^m$ for all $m$, and since a map $f\colon U\subseteq M \to \mathbb R^m$ is smooth if and only if $\ell\circ f\colon U \to \mathbb R$ is smooth for all linear functionals $\ell \colon \mathbb R^m\to \mathbb R$. It follows that (Yoneda) smooth structure of $M$ is equivalent to the data of the sheaf of smooth real-valued functions $\mathcal C^{\infty}(M)$ on $M$.
Thus it seems at least as useful to define $\mathcal C^{\infty}_M$, the sheaf of smooth functions on $M$ (which is fine, so you don't need to delve into sheaf cohomology). Why not call $\mathcal C^{\infty}_M$ the "smooth structure" for example? (To be a little bit more precise, one could say that $M$ is a (second countable etc.) topological space equipped with a (sheaf of) rings $\mathcal C^{\infty}_M$ with $\mathcal C^{\infty}_M(U)$ vector space of functions $f\colon U \to \mathbb R$, such that for any $x \in M$ there is an open neighbourhood $U_x$ of $x$ and a homeomorphism $\phi\colon U_x \to V$, where $V\subseteq \mathbb R^n$ is an open subset of $\mathbb R^n$ and $\phi$ is such that $\ell\circ \phi \in \mathcal C^\infty(U_x)$ for any linear functional $\ell\colon \mathbb R^n\to \mathbb R$.
In summary, does the notion of a maximal atlas actually ever do any work for you, or is it simply a cheap way to resolve what one means by a "smooth structure"?
Addendum: My hunch is that "maximal atlas" is a concept only used in teaching introductory courses: much as the rational number $1/2$ is strictly speaking an equivalence class of pairs $\{(1,2),(2,4),(3,6),...\}$ one can give an atlas and then say "the structure is the maximal atlas containing it" but then in reality, one never thinks of the equivalence class (which is not to say one might chose to consider another representative).
Addendum II: Thanks for all the responses! I'm not sure I'm convinced yet that the notion of maximal atlas does more work than providing an answer to the question of what a differentiable structure is. I agree that one certainly wants to have more charts that the initial atlas in order to do calculations, but that just requires the notion of a compatible chart (which is really already in the definition of an atlas) and not a maximal atlas. Of course in principle, once you have the notion of a chart being compatible with a given atlas, then the existence of a maximal atlas is straight-forward, so I could be accused of sophistry I suppose.
To attempt to defend myself on that one, I think I can now formulate better my objection to the maximal atlas notion though: I think the definition of a smooth manifold is quite a lot to absorb when one first sees it (there is an open cover, each of whose elements requires a homeomorphism to a subset of $\mathbb R^n$, and these have to satisfy a compatibility condition which requires being teleported up into the manifold and back down again....).
Thus it seems like one is adding an additional layer of complication by setting up the notion of a maximal atlas. Given one actually uses the algebra of smooth functions on a manifold fairly quickly in the theory (I'm mainly thinking of partitions of unity here, which also allows you to ignore the sheaf-theoretic technology) I still wonder if it would not be simpler to define a "smooth structure" to be the space of smooth functions $\mathcal C^{\infty}(M)$ (no sheaves, just all smooth functions, using partitions of unity to show no local-to-global issues can arise).
I should also say that the fact that maximal atlases don't appear to be a notion people find particularly straight-forward to absorb may be because most of the time in mathematics, the notion of a [insert favourite mathematical widget here] is stable under intersections, leading to the notion of the widget "generated by" some set (e.g. span of vectors, subgroups generated by subsets, topologies for which certain functions are continuous etc. etc.). For atlases however, we use union, and thus the maximal atlas is, I suppose, the atlas "co-generated" by the initial covering atlas we use to specify a smooth structure.
Addendum III Just to clarify, I don't mean to suggest at all that charts are a bad thing! What I am saying is that, if a topological manifold $M$ is given a smooth atlas $\mathcal A = \{(U_i,\phi_i): i \in I\}$ where, for $i \in I$ the map $\phi_i\colon U_i \to V_i\subseteq \mathbb R^n$, then for any open set $U\subseteq M$ we may define $$ \mathcal C^{\infty}_M(U) = \{ f\colon U \to \mathbb R: f\circ \phi^{-1}\colon V_i \to \mathbb R \} $$ Then, as already mentioned, the collection of algebras $\mathcal C^{\infty}_M(U)$ (which form a sheaf, but it may not be so important to say that initially) determines the maximal atlas, because a chart $(U,\varphi)$ is compatible with $\mathcal A$ if and only if $\ell\circ \varphi\colon U \to \mathbb R$ lies in $\mathcal C^{\infty}_M(U)$ for all (or a basis of the space of) linear functionals $\ell\colon \mathbb R^n \to \mathbb R$.
Given that, it seems to me better to declare $\mathcal C^{\infty}_M$ to be the "smooth structure" not the maximal atlas. This is for a couple of reasons: First, you use functions on open sets of a smooth manifold constantly, so the definition of the smooth structure is an object one uses, rather than an esoteric one stated and then forgotten. Secondly, it is arguably more accurate -- one uses a smooth (or $\mathcal C^k$ or $\mathcal C^{\omega}$ atlas in order to transport the notion of a function being $\mathcal C^{k,\infty,\omega,\ldots}$ onto the manifold $M$ from the notion on open subsets of $\mathbb R^n$, but it is the notion of smooth function or analytic function that is controlling the nature of the object you get.