Solution:

For a pair $m, n$ with $\operatorname{gcd}(|m|,|n|) = d > 1$, $m^{2} + k m n + n^{2}$ is divisible by $d$, so that the set $A$ could also be the set of all integer multiples of $d$, which does not contain the element $1$ and is therefore different from $\mathbb{Z}$.

Now we consider numbers $m, n$ with $\operatorname{gcd}(|m|,|n|) = 1$. For these we also have $\operatorname{gcd}(m^{2}, n^{2}) = 1$. Hence it follows from the extended Euclidean algorithm or from Fermat's little theorem that there exist integers $r$ and $s$ with $r m^{2} + s n^{2} = 1$. Moreover, for $x = y = m$ it is clear that $(2 + k) m^{2} \in A$, i.e. every integer multiple of $m^{2}$ lies in $A$ – correspondingly, every integer multiple of $n^{2}$ lies in $A$. And for $k = 2$ it follows that for all $x, y \in A$ we also have $(x + y)^{2}$ in $A$. Hence $r m^{2}, s n^{2}$ and consequently $(r m^{2} + s n^{2})^{2} = 1^{2} = 1$ lie in $A$.