I think it is not possible for a uniformly discrete set $E$ in any ${\bf R}^d$ to be paradoxical, because one can create an invariant (or almost invariant) mean on such a set. Indeed, for any $\varepsilon>0$ and $C>0$, one can use the pigeonhole principle to find a large radius $R$ such that $|E \cap (B(0,R+C) \backslash B(0,R-C))| \leq \varepsilon |E \cap B(0,R)|$, where $|A|$ denotes the cardinality of $A$, because the uniformly discrete nature of $E$ forces $|E \cap B(0,R)|$ to grow at most polynomially. This makes the probability measure $\mu(A) := |A \cap E \cap B(0,R)| / |E \cap B(0,R)|$ approximately invariant (up to error $\varepsilon$) with respect to isometries $T$ that involve a translation by at most $C$, and which map $A$ to a subset of $E$, in the sense that $|\mu(TA)-\mu(A)| \leq \varepsilon$ for such sets. This should be inconsistent with any putative paradoxical decomposition if one chooses $C,\varepsilon$ appropriately with respect to this decomposition. (One could also create a genuinely invariant mean by sending $R \to \infty$ and taking an ultralimit or by using a suitable compactness theorem, although this does not seem necessary for this particular application.)

(There should also be some way to use the Tits alternative or something related to this alternative (e.g. Jordan's theorem, Gromov's theorem on groups of polynomial growth, or the Solovay-Kitaev argument) to show the stronger assertion that the semigroup of isometries that map a uniformly discrete set to a subset should be "virtually nilpotent" (or maybe even virtually abelian) in some sense, although I was not able to make any of these fancier tools work here (largely because one only has a semigroup of isometries to play with rather than a group), whereas the more elementary counting argument above seems to already suffice for the purposes of answering the stated question.)