= Solution
Use the standard coordinate generators of $\mathbb Z^d$. Its <word metric> is the $\ell^1$ metric, so its radius-$r$ ball consists of integer points satisfying $\sum_i|x_i|\le r$. For $d\ge1$ it contains the cube $|x_i|\le\lfloor r/d\rfloor$ and is contained in the cube $|x_i|\le r$. Thus
$$
(2\lfloor r/d\rfloor+1)^d\le\beta_{\mathbb Z^d}(r)\le(2r+1)^d,
$$
and its growth is bounded above and below by constant multiples of $r^d$ for large $r$.
If $\mathbb Z^n$ and $\mathbb Z^m$ are quasi-isometric, part (i) gives equivalent <growth functions of a finitely generated group>. For positive $n>m$, the required comparison $r^n\lesssim r^m$ would bound a positive constant times $r^n$ by $O(r^m+r)$, impossible as $r\to\infty$. Interchanging $n,m$ rules out $m>n$. Hence $n=m$. Conversely, for equal dimensions the identity is an isometry with standard generators; changing finite generators preserves <quasi-isometry> by Q2(ii). Therefore
$$
\boxed{\mathbb Z^n\text{ and }\mathbb Z^m\text{ are quasi-isometric if and only if }n=m.}
$$
If dimension zero is allowed, $\mathbb Z^0$ is bounded and cannot be quasi-isometric to an unbounded positive-dimensional lattice. This boundedness argument is needed because the additive linear term in $\sim_e$ makes constant and linear <growth functions of a finitely generated group> equivalent; growth equivalence alone would not distinguish dimensions zero and one.
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