The area of a triangle $\Delta = rs$, where $r$ is its inradius and $s$ is its semiperimeter. The area of the incircle is $\pi r^2$. We want to maximize the ratio of the circle's area to the triangle's area; namely, the ratio $$\begin{align}\frac{\pi r^2}{rs} &= \frac{\pi r}{s} \\[4pt] &\propto \frac{r}{s} \end{align}$$
From $rs = \Delta = \sqrt{s(s-a)(s-b)(s-c)}$ where $a$, $b$, $c$ are the sides of the triangle, we obtain $r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$ whereupon
$$\frac{r}{s} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s^3}}$$
Since the quantity under the sqaure root is positive, maximizing $\frac{r}{s}$ is equivalent to maximizing $\sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}}$. By the AM-GM inequality,
$$\frac{(s-a) + (s-b) + (s-c)}{3} \geq \sqrt[3]{(s-a)(s-b)(s-c)} \\
\implies \frac{(s-a) + (s-b) + (s-c)}{3s} \geq \sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}}$$
but the left-hand side is merely $\frac{1}{3}$, so
$$\sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}} \leq \frac{1}{3}$$
Note that equality occurs only when $s - a = s - b = s - c \implies a = b = c$, or in other words, for an equilateral triangle. Thus, only when the triangle is equilateral, the maximum value of $\sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}}$ and thereby of $\frac{r}{s}$ and $\frac{\pi r^2}{rs}$ is obtained.
(Extra: the actual value of $\frac{\pi r^2}{rs}$ is slightly greater than $0.6$ for an equilateral triangle. The incircle covers slightly more than $60\%$ of the triangle's area.)