Let the ellipse be in standard position, with (fraction-free) equation
$$b^2 x^2 + a^2 y^2 = a^2 b^2$$
Let our equilateral triangle have circumcenter $(p,q)$ and circumradius $r$. Note that maximizing the area of the triangle is equivalent to maximizing $r$.
For some angle $\theta$ ---actually, for three choices of $\theta$--- the vertices of the triangle have coordinates
$$
(p,q) + r \; \mathrm{cis}\theta \qquad (p,q)+r\;\mathrm{cis}\left(\theta+120^{\circ}\right) \qquad (p,q) + r\;\mathrm{cis}\left(\theta-120^{\circ}\right)
$$
where I abuse the notation "$\mathrm{cis}\theta$" to indicate the vector $(\cos\theta,\sin\theta)$.
Substituting these coordinates into the ellipse equation gives a system of three equations in four parameters $p$, $q$, $r$, $\theta$. I used Mathematica's Resultant[] function to help me eliminate $r$ and $\theta$, arriving at a huge polynomial equation in $p$ and $q$. One factor of the polynomial gives rise to this equation:
$$p^2 b^2\left( a^2+3b^2 \right)^2 + q^2 a^2\left(3a^2+b^2\right)^2= a^2b^2\left(a^2-b^2\right)^2$$
This says that the family of circumcenters $(p,q)$ lie on their own ellipse! We can therefore write
$$p = \frac{a\left(a^2-b^2\right)}{a^2+3b^2}\cos\phi \qquad q = \frac{b\left(a^2-b^2\right)}{3a^2+b^2}\sin\phi$$
for some $\phi$. Back-substituting into the system of equations gives this formula for $r$:
$$r = \frac{4 a b \sqrt{a^2\left(a^2+3b^2\right)^2-\left(a-b\right)^3\left(a+b\right)^3 \cos^2\phi}}{\left(a^2 + 3 b^2\right)\left(3a^2+b^2\right)}$$
The maximum value, $R$, is attained when $\cos\phi = 0$, so
$$R := \frac{4a^2b}{3a^2+b^2}$$
The area of the maximal triangle is
$$\frac{3\sqrt{3}}{4}R^2 = \frac{12a^4 b^2\sqrt{3}}{\left(3a^2+b^2\right)^2}$$
Perhaps-unsurprisingly, the corresponding triangles are centered horizontally within the ellipse, with a vertex at either the top or bottom of the minor axis.
Now, I should point out that my big $pq$ polynomial has other factors, namely, $p$ itself, $q$ itself, and a giant I'll call $f$.
One can verify that the cases $p=0$ and $q=0$ lead to the same results as above. (Specifically, they correspond to the respective cases $\cos\phi=0$ and $\sin\phi=0$.) Intuitively, if a circle's center lies on an axis of the ellipse, then the points of intersection with the ellipse have reflective symmetry over that axis. If there are four distinct points (or two, or none), then we cannot choose three to be the vertices of our equilateral triangle; consequently, there must be only three points of intersection, with one of them on the axis, serving as the point of tangency for the circle and ellipse.
As for the case $f=0$ ... I'll just irresponsibly call it extraneous. (The method of resultants tends to spawn such things.)
Let $a=BC$, $b=AC$, $c=AB$ be the side lengths in the given triangle. Since $c^2=a^2+b^2$ we need two equations in $a,b$ to solve the issue.
$(1)$ First of all, there are similar triangles, the given one with proportion $CB:CA=a:b$, the one with corresponding proportion $(a-21):21$, and the one with corresponding proportion $21:(b-21)$, so
$$
\frac ab = \frac{a-21}{21}= \frac{21}{b-21} \ .
$$
Equivalently,
$$
ab = 21(a+b)\ .
$$
$(2)$ Consider now the other square, it has size $s=\sqrt{440}$. We have again two similar triangles in the picture, both have in $C$ the right angle, the given one with hypotenuse $c$ and height $h=ab/c$, and its "cut version" with hypotenuse $s$ and height $h-s$, so we obtain a second equation:
$$
\frac {h-s}s = \frac hc\ .
$$
Equivalently, $hc-sc=hs$, i.e. $hc^2=sc^2+hcs$, i.e. $abc=s(a^2+b^2+ab)$, and after squaring:
$$
a^2b^2(a^2+b^2) = 440(a^2+b^2+ab)^2\ .
$$
We solve the system involving the two equations, let $S$ be the sum $S=a+b$, the number asked for in the OP, and $P$ the product $P=ab$. Then we have:
$$
\left\{
\begin{aligned}
P &= 21 S\ ,
\\
P^2(S^2-2P) &= 440(S^2-P)^2\ .
\end{aligned}
\right.
$$
We plug in $P$ from the first equation in the second equation, get:
$$
(S + 420)(S - 462)S^2\ .
$$
We have thus $\color{blue}{S=462}$, since
we must reject for geometrical reasons the vanishing of the other factors.
To obtain $a,b$ explicitly, we have to solve the equation $x^2-Sx+P=0$, which is $x^2 - 462x + 21\cdot 462=0$. And indeed, the solutions $231\pm 63\sqrt{11}$ are real positive numbers, the two values of $a,b$.
As a final note, the given numbers are so ugly, that checking is done hard, hope there is no computational error in between.
Best Answer
Thanks to the hint of markvs, we can simplify the problem as follows:
Stretching the area back to the ellipse (along the x-axis) leads to the area:
$$T_{\triangle ABC}=\frac{a}{b}\cdot\frac{3}{4}\sqrt{3}b^2$$