Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $\vert y-x\vert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, {x+y\over 2}, {x+y\over 2}$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$\left\vert {y\over 2}-{x\over 2}\right\vert={1\over 2}\vert y-x\vert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $\vert y-x\vert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, ${1\over 2}$): if $r\in(-1,1)$ then for any $a$ we have
$$\lim_{n\rightarrow\infty}ar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
I'll use $\triangle PQR$ instead of $\triangle ABC$, to avoid some notational confusion.
First, a little prep work.
Given $\triangle PQR$, we erect on (directed) segment $\overline{PQ}$ an equilateral triangle $\triangle PQR'$ with a clockwise orientation. Similarly, we erect clockwise equilaterals $\triangle QRP'$ and $\triangle RPQ'$. As it happens, the lines $\overleftrightarrow{PP'}$, $\overleftrightarrow{QQ'}$, $\overleftrightarrow{RR'}$ meet at a common point, and do so symmetrically. When $\triangle PQR$ itself has a counter-clockwise orientation, the three equilaterals are external to it, and the three lines meet at the first isogonic center (Kimberling center $X_{13}$); when $\triangle PQR$ has a clockwise orientation, the equilaterals overlap its interior, and the common point is the second isogonic center ($X_{14}$).
In any case, we find that any triangle can be positioned such that each of its vertices lies on one of three concurrent, symmetrically-arranged lines. Taking the common point to be the origin, and one of the lines to be the $x$-axis, we can coordinatize $\triangle PQR$ thusly (abusing notation so that $\operatorname{cis}\theta := (\cos\theta, \sin\theta)$) as
$$P := p \operatorname{cis} 0 \qquad Q := q \operatorname{cis}\tfrac23\pi\qquad R := r \operatorname{cis}(-\tfrac23\pi) \tag{1}$$
where we may assume $q$ and $r$ are non-negative (and not simultaneously zero). One can show that the $x$-intercept of $\overline{QR}$ is $-qr/(q+r)$; consequently, $\triangle PQR$'s orientation depends upon $p$'s relation to that value, and we can write
$$\text{The origin is}\;\triangle PQR\text{'s}\; \left\{\begin{array}{c}\text{first} \\ \text{second} \\ \text{(either)} \end{array}\right\}\; \text{isogonic center if} \;\; p q + q r +r p \left\{\begin{array}{c} > \\ < \\ = \end{array}\right\} 0 \tag{2}$$
Now to the topic at hand.
The sides of an equilateral triangle circumscribing $\triangle PQR$ are lines through $P$, $Q$, $R$ with symmetrically-arranged normal vectors, say, $$u := \operatorname{cis}\theta \qquad v := \operatorname{cis}\left(\theta+\tfrac23\pi\right)
\qquad w := \operatorname{cis}\left(\theta-\tfrac23\pi\right) \tag{3}$$
Let $T_\theta$ be the resulting equilateral triangle. I'll forego giving its vertices. The important thing is to compare the sizes of these triangles across all $\theta$, which we can do by computing the areas:
$$|T_\theta| = \frac{1}{\sqrt{3}}\;\left(\,p + q + r\,\right)^2\,\cos^2\theta \tag{4}$$
Clearly, $|T_\theta|$ is maximized with $\theta = 0$ (or $\pi$), giving a maximal area
$$|T_0| = \frac1{\sqrt{3}} (p+q+r)^2 \tag{$\star$}$$
when $u$, $v$, $w$ are parallel to $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$. When $O$ is $\triangle PQR$'s first isogonic center (see $(2)$), then $T_0$ matches OP's construction.
Now, while the $T_\theta$ triangles form an infinite family of equilaterals with side-lines containing the vertices of $\triangle PQR$, theirs is not the only such family. We get another by changing the signs in $(3)$; that is, by exchanging the roles of $v$ and $w$. The corresponding triangles, $T_\theta'$, have areas given by
$$|T^\prime_\theta| = \frac{1}{\sqrt{3}}\,\left(\,p \cos\theta + q \cos\left(\theta+\tfrac23\pi\right) + r \cos\left(\theta-\tfrac23\pi\right) \,\right)^2 \tag{5}$$
A quick derivative tells us that the critical values of $(4)$ occur for
$$\cot\theta = \frac{\sqrt{3}\,(q-r)}{2p-q-r} \quad\text{or}\quad
\tan\theta = \frac{\sqrt{3}\,(r-q)}{2p-q-r} \tag{6}$$
The former corresponds to an area of zero; the latter is maximizing, and we have
$$\phi := \tan^{-1}\frac{\sqrt{3}\,(r-q)}{2p-q-r} \quad\to\quad |T^\prime_\phi| = \frac1{\sqrt{3}} \left(p^2+q^2+r^2-p q-q r-r p\right) \tag{$\star\star$}$$
Comparing $(\star)$ to $(\star\star)$, we have
$$|T_0| - |T_\phi^\prime| = \sqrt{3} \left(p q + q r + r p\right) \tag{7}$$
which hearkens back to $(2)$. So, $T_0$ is maximal when $O$ is the first isogonic center of $\triangle PQR$; otherwise, $T^\prime_\phi$ is.
Importantly, the reader can verify that, at the key angle $\theta = \phi$ from $(\star\star)$, the lines through $P$, $Q$, $R$ with the direction vectors $u$, $v$, $w$ (the last two with their signs exchanged) meet at a point; specifically, they meet at $\triangle PQR$'s "other" isogonic center. This tells us that $T^\prime_\phi$ is actually the equilateral triangle obtained from OP's construction relative to that "other" center. Since, by $(7)$ and $(2)$, equilateral $T^\prime_\phi$ maximizes all triangles precisely when the "other" center is the first isogonic center, we have shown that
OP's construction relative to the first isogonic center is always maximal.
$\square$
Here are a couple of animations, showing differently-oriented $\triangle PQR$ (counter-clockwise vs clockwise); the origin (unmarked black dot) is either the first or second isogonic center, respectively. Triangles $T_\theta$ are green, while $T^\prime_\theta$ are light blue.
These images highlight that, while the various equilateral triangles have side-lines passing through the vertices of $\triangle PQR$, not all have $\triangle PQR$ in their interiors; hence, they are not all "circumscribing" in the traditional sense.
Best Answer
For the record, I will solve the much easier problem of finding infinitely many rational points on the circumcircle. Let's consider the case of a point $P$ on the arc from $A$ to $B$. By Ptolemy's theorem, $|PA| + |PB| = |PC|$, so if $|PA|$ and $|PB|$ are rational then so is $|PC|$. Also, $P$ is on the arc from $A$ to $B$ if and only if $P$ is on the right side of $\overline{AB}$ and $\angle APB = 120^{\circ}$. By the Law of Cosines, $\angle APB = 120^{\circ}$ is equivalent to $$|PA|^2 + |PA| |PB| +|PB|^2=1.$$
This conic can be paramterized in the usual way: Put $|PA|=1+t$, $|PB|=kt$. Solve for $t$ in terms of $k$; the result is $t=-(k+2)/(k^2+k+1)$. So $$|PA| = \frac{k^2-1}{k^2+k+1} \quad |PB| = \frac{-k^2-2k}{k^2+k+1} \quad |PC|=\frac{-2k-1}{k^2+k+1}$$ or, in other words, $$a=k^2-1 \quad b= -k^2-2k \quad c=-2k-1 \quad d=k^2+k+1.$$ We want to have $-2 < k < -1$ to get the right signs.
I don't want the bounty for this though; I want mathlove to explain how the heck he or she found his or her solution.
I have figured out a way to get mathlove's answer. I'll write $PA=a/d$, $PB=b/d$ and $PC=c/d$. As described here, these obey the relation $$a^4+b^4+c^4+d^4 = a^2 b^2 + a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 + c^2 d^2.$$ Let $\Sigma$ be the surface in $\mathbb{P}^3$ cut out by this degree $4$ equation. Notice that $\Sigma$ has $16$ singular points: The $4$ points $(\pm 1 : \pm 1 : \pm 1 : 0)$ and the other $12$ which come from putting the zero in the other possible positions. The three points $(1:1:0:1)$, $(1:0:1:1)$ and $(0:1:1:1)$ are the vertices of the triangle; the other $13$ singularities involve negative or infinite values for $(PA, PB, PC)$.
The technical term for this is a Kummer surface and, in fact, $\Sigma$ is the special kind of Kummer surface called a tetrahedroid. But we don't need to know this to follow the rest of the argument.
Take a plane through any three of the singularities. The resulting planar slice of $\Sigma$ will be a degree $4$ plane curve with $\geq 3$ nodes. If there are exactly $3$ nodes, the resulting curve is genus $0$, and thus has a rational parametrization over $\mathbb{C}$. That parmetrization doesn't have to have rational coefficients but, in some lucky cases, it does. We also aren't promised that the resulting values of $(PA, PB,PC)$ will be positive, let alone inside the triangle but, again, sometimes we get luck.
Mathlove uses the plane $b+d=2a$, passing through the points $(0,1,1,-1)$, $(0,-1,1,1)$ and $(1,1,0,1)$. The complete list of planes, up to permuting $(a,b,c,d)$ and switching signs, is $a=0$, $a+b=0$, $a+b+c=0$, $a+b+2c=0$ and $a+b+2c+3d=0$. These symmetries of $\Sigma$ do not respect the condition that the points actually correspond to physical points inside the triangle, so you have to keep track of more possibilities if you want that to hold.
Several of these planes correspond to interesting geometric configurations:
The equations $a+d=b$, $a+b=d$ and $b+d=a$ are $PA+1=PB$, $PA+PB=1$ and $PB+1=PA$ respectively, which say that $P$ lies on the line $AB$ (either in the two unbounded rays or in the line segment $AB$.)
The equation $a+b=c$ means $PA+PB=PC$, so (by Ptolemy's theorem) $P$ is on the arc of the circumcircle from $A$ to $B$; the other arcs of the circumcircle are described similarly.
The equation $a=b$ means that $P$ is on the perpendicular bisector of $AB$. This doesn't actually contribute any points; the intersection of $\Sigma$ with $\{ a=b \}$ the product of two conics, neither of which has rational coefficients.
Other planes, such as mathlove's choice $PB+1=2PA$ have no clear geometric meaning, but we can still rationally parametrize them and, at least in some cases, it seems we win.
As far as I can tell from skimming papers, there is an enormous literature on rational points on Kummer surfaces, but there isn't one simple answer. I had hoped to use this question as an opportunity to teach myself about Kummer surfaces (and, to some extent, I have) but it looks it's a big field, so I'll stop here.