A morphism of schemes $f : X \to Y$ is smooth iff it it is locally of finite presentation and it is formally smooth; the latter means that for every affine scheme $\mathrm{Spec}(R)$ over $Y$ and every ideal $I \subseteq R$ with $I^2=0$ the canonical map $\hom_Y(\mathrm{Spec}(R),X) \to \hom_Y(\mathrm{Spec}(R/I),X)$ is surjective. Geometrically this means that every point can be thickened to an infinitesimal point. For a proof, see EGA IV, 17.
In particular, $A \to B$ is a smooth ring homomorphism iff it is of finite type and for every $A$-algebra $R$ and every ideal $I \subseteq R$ with $I^2=0$ the canonical map $\hom_A(B,R) \to \hom_A(B,R/I)$ is surjective. For $\mathbb{Z} \to \mathbb{Z}[T,T^{-1}]$ this map corresponds to $R^* \to (R/I)^*$. And yes, it is surjective (elementary calculation using that $I$ is nilpotent).
Of course, there are other ways to see that this is smooth. Every affine n-space $\mathbb{A}^n_S$ is smooth over $S$ (follows directly from the Jacobi definition or the characterization above). Also every open immersion is a smooth (even étale) morphism. Smooth morphisms are closed under composition. Thus every open subscheme of $\mathbb{A}^n_S$ is smooth over $S$. In particular, $\mathrm{Spec}(\mathbb{Z}[T,T^{-1}])$ is smooth.
As for your other question: Smoothness can be checked locally on the base. Therefore you may assume that your vector bundle is trivial, thereby reducing to the smoothness of the affine n-space.