Let $(R, +, \times, 0, 1)$ be a ring where $0$ is the identity of $+$ and $1$ is the identity of $\times$.
A regular element of a R is an element such that it is neither a left-divisor, or a right divisor of $0$. That is, suppose $a$ is a regular element of $R$, then for any $x \in R$: $$ a \times x \neq 0$$ $$ x \times a \neq 0$$
An ideal $I$ is such that $(I, +)$ is a sub-group of $(R, +)$, and for any $x \in R$, if $y \in I$, then $y \times x \in Y$. A proper ideal is an ideal $I$ of $R$ such that $I \neq R$.
I read the following in an article I am trying to understand:
It seems that I want to show that if $a$ is regular, then $a \not\in I$, where $I$ is a proper ideal?
I am using Wikipedia as the source of my definitions, so what I have written about is my (mis)interpretation of Wikipedia's definitions.
You're quoting a different definition of “regular” than used in the paper. Note that $^*\mathbb{R}$ is a field, so there is no zero-divisor and the definition you quote applies to every element except $0$.
For the cited paper, “regular” is synonymous with “invertible”. The set $\vartheta$ consisting of the infinitesimal numbers is precisely the set of noninvertible elements in $\mathcal{O}$, because their inverse in $^*\mathbb{R}$ is infinite and $\mathcal{O}$ just consists of the finite numbers.
The paragraph you got stuck with simply says that $\vartheta$ is a maximal ideal because $\mathcal{O}/\vartheta$ is isomorphic to $\mathbb{R}$, which is a field.