Non-unital ring of rational numbers with even numerator and odd denominator has no maximal ideals

This is false. Let $R=\{\frac{2n}{2m+1}:n,m\in\mathbb{Z}\}$ and let $I=2R=\{\frac{4n}{2m+1}:n,m\in\mathbb{Z}\}$, which is a proper ideal. Note that $rs\in I$ for all $r,s\in R$, so the quotient $R/I$ is a rng in which all products are $0$. This means ideals in $R/I$ are the same as additive subgroups. As an abelian group, $R/I$ is a vector space over $\mathbb{Z}/(2)$ since every element is annihilated by $2$. So, we can pick a codimension $1$ vector subspace $J\subset R/I$ and $J$ will be a maximal proper subgroup and hence a maximal ideal. The inverse image of $J$ in $R$ is then a maximal ideal in $R$.

This looks like an attempt to use the description found in the final paragraph of this paper linked to in the comments. It reads:

One can create more rings satisfying the hypothesis of the corollary by starting with any commutative ring with identify and divisible additive group, taking its localization $S_M$ at a maximal ideal $M$, and letting $R=MS_M$.

The corollary in question says, inter alia, that if $S$ is a commutative ring with identity and a unique maximal ideal $R$, and its additive group is divisible, then $R$ has no maximal ideals. Alternatively, that if $R^2+pR=R$ for every integer prime $p$, then $R$ has no maximal ideals.

It looks to me like perhaps this is an attempt at doing that construction by taking the ring of rationals with odd denominator; that is, the localization of $\mathbb{Z}$ at $(2)$, as the ring, and then taking $(2)\mathbb{Z}_{(2)}$. The problem is that this ring does not satisfy the hypotheses of the corollary: the underlying additive group of $\mathbb{Z}_{(2)}$ is not divisible, and $(2)^2+2(2)\neq(2)$ in that ring. So this is not an instance of the given construction. You can also not start with $\mathbb{Q}$ itself, because then the maximal ideal is $(0)$ and you just get the trivial ring.