Edit: I have seen the proof involving rational areas, but I came at this problem trying to use the roots of unity and rotations and translations of the complex plane.
I couldn’t find a proof along these lines elsewhere, so posting here for verification.
Proof: Assume WLOG that $\triangle$ is an equilateral triangle with vertices $z_1=(0,0), z_2=(a,b), z_3=(c,d)$ on the lattice.
Let $\omega=e^{\frac{2\pi i}{3}}$ be a cube root of unity, then it is easy to show$$\tag{1} 1+\omega + \omega^2=0 \implies 1+\omega^2=-\omega$$ and $$\tag{2} -\omega^2=e^{\frac{\pi i}{3}}.$$ Since $ \triangle$ is equilateral, say a rotation about $z_2$ through $\frac{\pi}{3}$ radians takes $z_3$ to $z_1$ (the result is unchanged if the vertices are swapped), from which it follows $$ z_1-z_2= -\omega^2 (z_3-z_2)$$ $$\implies z_1-(1+\omega^2)z_2+\omega^2z_3=0$$ $$\implies z_1 + \omega z_2 + \omega^2 z_3 = 0 \tag{3}.$$
From $(3)$ and the fact that $z_1=(0,0)$ we get$$\tag{4} \omega z_2=-\omega^2z_3= e^{\frac{\pi i}{3}}z_3 \implies z_2= e^{\frac{5\pi i}{3}}z_3,$$ but, converting from polar form and equating real and imaginary parts, this implies $$ \tag{5} a= \frac{c}{2} + \frac{d}{2}\sqrt{3}$$ and $$\tag{6} b=\frac{d}{2} - \frac{c}{2}\sqrt{3}.$$ Since $c$ and $d$ cannot both be $0$, we have a contradiction. $\square$