It doesn't matter how many unknowns there are. The polynomial ring is still spanned by monomials consisting of a finite number of the variables at a time, so multi-indices work fine and so does the degree.
Formally, as hinted at in the comments, the polynomial ring $k[S]$ on a set $S$ over a commutative ring $k$ is the monoid algebra over $k$ on the direct sum $\bigoplus_S \mathbb{Z}_{\ge 0}$ of $S$ many copies of $\mathbb{Z}_{\ge 0}$, or in other words, the free commutative monoid on $S$. (Not the direct product $\mathbb{Z}_{\ge 0}^S$, that's much larger.)
Abstractly, if we define the polynomial $k$-algebra functor $k[S]$ as the left adjoint to the forgetful functor from commutative $k$-algebras to sets, the forgetful functor factors as a composite
$$\text{CAlg}(k) \xrightarrow{U(-)} \text{CMon} \xrightarrow{U'(-)} \text{Set}$$
through commutative monoids, where $U$ is given by forgetting everything except the multiplication, and so its left adjoint factors as a composite going the other way ("adjoints compose")
$$\text{Set} \xrightarrow{F'(-)} \text{CMon} \xrightarrow{F(-)} \text{CAlg}(k)$$
where $F'(-)$ takes the free commutative monoid on a set and $F(-)$ takes the monoid $k$-algebra of a commutative monoid.