I found the same problem on page 13 in Richard Courant's classic textbook Differential and Integral Calculus:
2.* In an ordinary system of rectangular co-ordinates, the points for which both co-ordinates are integers are called lattice points. Prove that a triangle whose vertices are lattice points cannot be equilateral
My proposed solution is to set the origin at a vertex. This can be done w.l.o.g. For every lattice point $P : (p,q)$ we get an equilateral triangle with vertex $P$ and a third vertex $X$ (we also get another triangle with $P$ and $X'$, see figure) where both vertices lie on the circumference of a circle with radius $r =\sqrt(p^2+q^2)$. The angle of the $OP$ edge is $\alpha$, and $\sin a = \frac q r, \ \cos \alpha = \frac p r$.
Figure: Two equilateral triangles for every lattice point $P$
We get the coordinates of $X$ on the circle (and in a similar way for $X'$):
$(x,y) = (r \cos(\alpha+\frac \pi 3), r \sin(\alpha + \frac \pi 3))$
The addition formulas for sine and cosine give,
$(x,y) = \left(r \left[ \cos \alpha \cos \frac \pi 3 - \sin \alpha \sin \frac \pi 3 \right], r \left[ \sin \alpha \cos \frac \pi 3 + \cos \alpha \sin \frac \pi 3 \right] \right) = \left( r \left[ \frac p {2r} -\frac {q \sqrt 3} {2r} \right], r \left[ \frac q {2r} + \frac {p \sqrt 3} {2r} \right] \right) = \left( \frac p 2 - \frac{ \sqrt 3 q} 2, \frac q 2 + \frac{\sqrt 3 p} 2 \right)$
If $ p,q \in \Bbb Z$ then $X$ is not a lattice point.