A pair of integers $m, n$ fulfills the condition if and only if $\gcd(m, n) = 1$. Suppose that $\gcd(m, n) = d > 1$. The set
$$A = \{ \ldots, -2d, -d, 0, d, 2d, \ldots \}$$
is admissible, because if $d$ divides $x$ and $y$ then it divides $x^{2} + k x y + y^{2}$ for every integer $k$. Also $m, n \in A$ and $A \neq \mathbb{Z}$.

Now let $\gcd(m, n) = 1$, and let $A$ be an admissible set containing $m$ and $n$. We use the following observations to prove that $A = \mathbb{Z}$:

(i) $k x^{2} \in A$ for every $x \in A$ and every integer $k$.

(ii) $(x + y)^{2} \in A$ for all $x, y \in A$.

To justify (i) let $y = x$ in the definition of an admissible set; to justify (ii) let $k = 2$.

Since $\gcd(m, n) = 1$, we also have $\gcd(m^{2}, n^{2}) = 1$. Hence one can find integers $a, b$ such that $a m^{2} + b n^{2} = 1$. It follows from (i) that $a m^{2} \in A$ and $b n^{2} \in A$. Now we deduce from (ii) that $1 = (a m^{2} + b n^{2})^{2} \in A$. But if $1 \in A$ then (i) implies $k \in A$ for every integer $k$.