$\newcommand{\de}{\delta}\newcommand{\M}{\mathcal M}
\newcommand{\ep}{\varepsilon}
\newcommand{\thh}{\theta}\newcommand\I{\mathcal I}\newcommand{\Si}{\Sigma}$Here is the (slightly edited) definition of the mutual information given in the answer linked in the OP:

Let $D$ be any discrete random variable (r.v.) with distinct values $d_i$ taken with probabilities $p_i=P(D=d_i)>0$ for $i\in I$, where $I$ is a denumerable (that is, at most countable) set. Let $X$ be any r.v. (defined on the same probability space as $D$), with values in any nonempty set $S$ (given also some sigma-algebra $\Sigma$ over $S$, to make $S$ a measurable space). Let $\nu$ be the probability distribution of $X$, so that $\nu(B)=P(X\in B)$ for all $B\in\Sigma$. For each $i\in I$ and each $B\in\Sigma$, let
\begin{equation*}
\nu_i(B):=P(D=d_i,X\in B).
\end{equation*}
Then $\nu_i$ is a (sub-probability) measure absolutely continuous with respect to $\nu$, so that we can consider a Radon--Nikodym density
\begin{equation*}
\rho_i:=\frac{d\nu_i}{d\nu}
\end{equation*}
of the measure $\nu_i$ with respect to $\nu$, so that the values of $\rho_i$ are in $[0,1]$.

Then the mutual information between $D$ and $X$ is defined as follows:
\begin{equation*}
\I(D,X):=\sum_{i\in I}\int_S d\nu\;\rho_i\ln\frac{\rho_i}{p_i}.
\end{equation*}

Let us now answer the current question.

Much more generally than in the OP, let $X$ be any r.v. as in the highlighted text above. Let $\nu$ denote the distribution of $X$.

Let $\M$ denote the set of all Bernoulli r.v.'s $Y$ (defined on the same probability space as $X$) with $P(Y=1)\in(0,1)$;
we suppose that the probability space is rich enough so that for any coupling $\gamma$ of any Bernoulli distribution and the distribution $\nu$ of $X$ there is a r.v. $Y$ such that the joint distribution of the pair $(Y,X)$ is $\gamma$.

In view of the highlighted definition, for any $Y\in\M$,
\begin{equation*}
\I(Y,X)=\int_S d\nu\;
\Big(\rho\ln\frac{\rho}{p}+(1-\rho)\ln\frac{1-\rho}{1-p}\Big),
\end{equation*}
where
$p:=P(Y=1)$, $\rho:=\frac{d\nu_1}{d\nu}$, and $\nu_1(B):=P(Y=1,X\in B)$ for all $B\in\Si$; we are assuming the convention that $0\times\text{anything}:=0$.

We want to maximize $\I(Y,X)$ over all $Y\in\M$. This is very easy to do, noting that $u\ln\frac{u}{p}+(1-u)\ln\frac{1-u}{1-p}$ is convex in $u\in[0,1]$. So, $\I(Y,X)$ is maximized when $\rho=1_A$ for some $A\in\Si$, and then $p=\nu(A)$ and $\I(Y,X)=\int_A d\nu\;
\ln\frac1{p}=p\ln\frac1p$. Maximizing the latter expression in $p\in(0,1)$, we see that, contrary to the conjecture in the OP,

\begin{equation}
\max_{Y\in\M}\I(Y,X)=\frac1e,
\end{equation}
for any r.v. $X$ whatsoever provided that $X$ is defined on a rich enough probability space.

jointdistribution. $\endgroup$