Problem: Find all real numbers $c > 0$ such that there exists a labeling
of the lattice points in $\ZZ^2$ with positive integers for which:
\begin{itemize}
  \ii only finitely many distinct labels occur, and
  \ii for each label $i$, the distance between any two points
  labeled $i$ is at least $c^i$.
\end{itemize}

Solution: The answer is $c < \sqrt2$. Here is a solution with Calvin Deng.

The construction for any $c < \sqrt 2$ can be done as follows.
Checkerboard color the lattice points and label the black ones with $1$.
The white points then form a copy of $\ZZ^2$ again
scaled up by $\sqrt 2$ so we can repeat the procedure with $2$
on half the resulting points.
Continue this dyadic construction until a large $N$
for which $c^N < 2^{\half (N-1)}$,
at which point we can just label all the points with $N$.

I'll now prove that $c = \sqrt 2$ (and hence $c \ge \sqrt 2$) can't be done.
\begin{claim*}
  It is impossible to fill a $2^n \times 2^n$ square
  with labels not exceeding $2n$.
\end{claim*}
The case $n = 1$ is clear.
So now assume it's true up to $n-1$;
and assume for contradiction a $2^n \times 2^n$ square
$S$ only contains labels up to $2n$.
(Of course every $2^{n-1} \times 2^{n-1}$ square
contains an instance of a label at least $2n-1$.)

\begin{center}
\begin{asy}
unitsize(0.75cm);

for (int i=0; i<=8; ++i) {
  if ((i!=0) && (i!=4) && (i!=8)) {
    draw( (0,i)--(8,i), gray );
    draw( (i,0)--(i,8), gray );
  }
  else {
    draw( (0,i)--(8,i), black+1 );
    draw( (i,0)--(i,8), black+1 );
  }
}


filldraw( box((2.1,4.1),(5.9,7.9)), opacity(0.2)+palered, red+1 );
filldraw( box((1.1,2.1),(4.9,5.9)), opacity(0.2)+lightgreen, blue+1 );
fill(box((0,4),(4,8)), opacity(0.1)+lightcyan);
label("$A$", (3.5,8), dir(90), red);
label("$B$", (2.5,2), dir(-90), blue);

label("$2$", (0.5,7.5));
label("$1$", (1.5,7.5));
label("$2$", (2.5,7.5));
label("$1$", (3.5,7.5));
label("$1$", (0.5,6.5));
label("$5$", (1.5,6.5), red);
label("$1$", (2.5,6.5));
label("$3$", (3.5,6.5));
label("$2$", (0.5,5.5));
label("$1$", (1.5,5.5));
label("$2$", (2.5,5.5));
label("$1$", (3.5,5.5));
label("$1$", (0.5,4.5));
label("$3$", (1.5,4.5));
label("$1$", (2.5,4.5));
label("$4$", (3.5,4.5));
label("$6$", (5.5,6.5), blue);
\end{asy}
\end{center}

Now, we contend there are fewer than four copies of $2n$:
\begin{lemma*}
  In a unit square, among any four points,
  two of these points have distance $\le 1$ apart.
\end{lemma*}
\begin{proof}
  Look at the four rays emanating from the origin and note
  that two of them have included angle $\le 90\dg$.
\end{proof}

So WLOG the northwest quadrant has no $2n$'s.
Take a $2n-1$ in the northwest and draw a square of size $2^{n-1} \times 2^{n-1}$
directly right of it (with its top edge coinciding with the top of $S$).
Then $A$ can't contain $2n-1$, so it must contain a $2n$ label;
that $2n$ label must be in the northeast quadrant.

Then we define a square $B$ of size $2^{n-1} \times 2^{n-1}$ as follows.
If $2n-1$ is at least as high $2n$, let $B$ be a $2^{n-1} \times 2^{n-1}$
square which touches $2n-1$ north and is bounded east by $2n$.
Otherwise let $B$ be the square that touches $2n-1$ west
and is bounded north by $2n$.
We then observe $B$ can neither have $2n-1$ nor $2n$ in it, contradiction.

\begin{remark*}
  To my knowledge, essentially all density arguments fail
  because of hexagonal lattice packing.
\end{remark*}