If the map from $X$ to the maxspec of $\mathcal O(X)$ is a bijection, then $X$ is indeed affine.
Here is an argument:
By assumption $X \to $ maxspec $\mathcal O(X)$ is bijective, thus quasi-finite,
and so by (Grothendieck's form of) Zariski's main theorem, this map factors as an open embedding of $X$ into a variety that is finite over maxspec $\mathcal O(X)$.
Any variety finite over an affine variety is again affine, and hence $X$ is an open subset of an affine variety, i.e. quasi-affine. So we are reduced to considering the case when $X$ is quasi-affine, which is well-known and straightforward.
(I'm not sure that the full strength of ZMT is needed, but it is a natural tool
to exploit to get mileage out of the assumption of a morphism having finite fibres, which is what your bijectivity hypothesis gives.)
In fact, the argument shows something stronger: suppose that we just assume
that the morphism $X \to $ maxspec $\mathcal O(X)$ has finite non-empty fibres,
i.e. is quasi-finite and surjective.
Then the same argument with ZMT shows that $X$ is quasi-affine. But it is standard that the map $X \to $ maxspec $\mathcal O(X)$ is an open immersion when $X$ is quasi-affine,
and since by assumption it is surjecive, it is an isomorphism.
Note that if we omit one of the hypotheses of surjectivity or quasi-finiteness, we can find a non-affine $X$ satisfying the other hypothesis.
E.g. if $X = \mathbb A^2 \setminus \{0\}$ (the basic example of a quasi-affine,
but non-affine, variety), then maxspec $\mathcal O(X) = \mathbb A^2$, and the open immersion $X \to \mathbb A^2$ is evidently not surjective.
E.g. if $X = \mathbb A^2$ blown up at $0$, then maxspec $\mathcal O(X) =
\mathbb A^2$, and $X \to \mathbb A^2$ is surjective, but has an infinite fibre
over $0$.
Caveat/correction: I should add the following caveat, namely that it is not always true, for a variety $X$ over a field $k$, that $\mathcal O(X)$ is finitely generated over $k$, in which case maxspec may not be such a good construction to apply, and the above argument may not go through. So in order to conclude that $X$ is affine, one should first insist that $\mathcal O(X)$ is finitely generated over $k$, and then that futhermore the natural map $X \to $ maxspec $\mathcal O(X)$ is quasi-finite and surjective.
(Of course, one could work more generally with arbitrary schemes and Spec rather than
maxspec, but I haven't thought about this general setting: in particular, ZMT requires some finiteness hypotheses, and I haven't thought about what conditions might guarantee that the map $X \to $ Spec $\mathcal O(X)$ satisfies them.)
Incidentally, for an example of a quasi-projective variety with non-finitely generated ring of regular functions, see this note of Ravi Vakil's
A family of open sets may not lead to a open set in the product. Take $k=\mathbb{C}$ and let $X=Y=\mathbb{A}^1_\mathbb{C}$ and let $$V_u = Y\setminus \{e^u\}$$ for each $u\in U=X$.
Clearly $V_u$ is open in $Y$.
However $P = \bigcup_{u\in X} \{u\} \times V_u$ is the complement in $\mathbb{A}^2_\mathbb{C}$ of the graph $\Gamma$ of the exponential function. Hence $P$ is open iff $\Gamma$ is closed (in the Zariski topology) and the later is not true because the exponential function is not algebraic.
(Note that it also gives an example of a set that is open in the euclidean topology but not in the Zariski topology.)
The, maybe only, reasonable way to define the Zariski topology of a quasiprojective variety is through an embedding. For instance if $X,Y$ are affine, $X\subset \mathbb{A}^n$, $Y\subset \mathbb{A}^m$, then the Zariski topology of $X\times Y$ is generated by $U\cap X\times Y$ where $U\subset \mathbb{A}^{n+m}$ is open.
Best Answer
Given any Noetherian topological space $(X,\tau)$, if you adjoin one more closed set $C$ the topology always remains Noetherian. That is, let $\tau'$ be the topology generated by $\tau\cup\{X\setminus C\}$; I claim $\tau'$ is Noetherian too. To simplify the argument, I will in fact enlarge $\tau'$ further to the topology $\tau''$ generated by $\tau\cup\{C,X\setminus C\}$. This topology $\tau''$ has a very simple description: it is just the disjoint union topology on $X$ when we decompose $X$ as $C\coprod (X\setminus C)$ and give both $C$ and $X\setminus C$ the subspace topology induced by $\tau$. But $C$ and $X\setminus C$ are both Noetherian in the subspace topology, and a disjoint union of finitely many Noetherian spaces is Noetherian, so $\tau''$ is Noetherian. Explicitly, given a descending sequence of $\tau''$-closed sets, their intersections with $C$ must eventually stabilize and their intersections with $X\setminus C$ must also eventually stabilize, so the entire sequence must eventually stabilize.
So, there do not exist any maximal Noetherian topologies, besides the discrete topologies on finite sets. Any non-discrete Noetherian topology can always be enlarged to another Noetherian topology.