Let the vertices of our triangle be $(0,0)$, $(a,b)$, and $(c,d)$, where $a$, $b$, $c$, and $d$ are integers. If all edge lengths are the same, then
$$a^2+b^2=c^2+d^2=(a-c)^2+(b-d)^2.$$
Minor manipulation turns this into
$$a^2+b^2=c^2+d^2=2ac+2bd.$$
Now we use my favourite identity, which was known more than a millenium ago in India, and even earlier by Diophantus, and so has often been called the Fermat Identity:
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.\qquad\qquad(\ast)$$
This identity can be easily verified by expanding both sides, or more conceptually by noting that the norm of the product of two complex numbers is the product of the norms.
Let $N=a^2+b^2=c^2+d^2=**2(ac+bd)**$. Then $ac+bd=N/2$. The identity $(\ast)$ now gives
$$N^2=\frac{N^2}{4}+(ad-bc)^2$$
or equivalently
$$3N^2=4(ad-bc)^2.$$
This is impossible, since $3$ times the perfect square $N^2$ cannot be a square unless $N=0$, which gives a very tiny triangle.
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.
Best Answer
No, because that would imply an infinite sequence of smaller and smaller triangles with the same property:
$\hspace{90pt}$
The key to the proof below is this property:
This implies that we can rotate points of $\mathbb{Z}^2$ around other points of $\mathbb{Z}^2$ by $\frac{\pi}{2}$ and we will still end up in $\mathbb{Z}^2$.
The proof:
Consider an equilateral triangle $\triangle ABC$ with vertices in $\mathbb{Z}^2$ and perform a rotation of $A$ around $C$ to get $A''$ which also has integer coordinates:
$\hspace{70pt}$
Then translate $B$ along $\vec{A''C}$ to get $B'$, again with integer coordinates:
$\hspace{70pt}$
Do this two more times to get also $A'$ and $C'$, all with integer coordinates:
$\hspace{80pt}$
However, observe that $\triangle A'B'C'$ is also an equilateral triangle with vertices in $\mathbb{Z}^2$, but it is strictly smaller than $\triangle ABC$ (for convenience marked with the gray shaded area). This implies an infinite descending chain of equilateral triangles with coordinates in $\mathbb{Z}^2$ which is clearly impossible.
Edit:
For those of you who would like a construction in which the relative sizes are more apparent, observe that existence of an equilateral triangle with vertices in $\mathbb{Z}^2$ implies existence of a regular hexagon with the same property, and that in turn implies an infinite sequence of smaller and smaller regular hexagons:
$\hspace{30pt}$
I once saw this method applied just for hexagons, but was unable to find the source (if someone knows it, I would be grateful for the reference).
I hope this helps $\ddot\smile$