This is an expansion of the comment of Qiaochu Yuan.

As mentioned in the comments there can be no "constructive" description. However, maybe you'll find this (tautological) construction useful: take any sequence $m_n$, $n=1,2,\ldots$ of real points which diverges to infinity. Choose an ultrafilter $U$, and define $M$ to be the set of those functions $f$ for which there exists an element $K$ of $U$ (so in particular $K$ is a subset of natural numbers) such that $f(m_k)=0$ for $k\in K$.

It's obvious that $M$ contains your set $I$ and that it is closed under multiplication by elements of $R$. That it is additively closed follows from the fact that $U$ is a filter, and that it is maximal follows from the fact that $U$ is an ultrafilter.

So $M$ is "as explicit" as $U$ is. In particular all functions which vanish on all but finitely many points of the sequence $m_n$ are in $M$.

Also, you can take your favourite explicit filter $U'$ and define the ideal $M'$ as above, and have a non-maximal but explicit ideal which contains $I$.