Let $p \equiv 3 \pmod 4$ be a prime number. We will show that for $k = p^2$ there are no positive integers $m$ and $n$ satisfying the equation. Suppose that there is a solution. If we rewrite the equation as
$$n^2 - p^2 n + (m^2 - p^2 m^4) = 0$$
we obtain a quadratic equation for $n$ and the discriminant must be a perfect square. Therefore,
$$p^4 + 4p^2 m^4 - 4m^2 = s^2$$
for some integer $s$. Then $p \mid (2m)^2 + s^2$ and we get $m = px$ and $s = pt$ for some integers $x$ and $t$. So we have
$$p^2 + 4p^4 x^4 - 4x^2 = t^2$$
and again we obtain $p \mid (2x)^2 + t^2$. Thus, we have $x = py$ and $t = pu$ for some integers $y$ and $u$. Dividing both sides by $p^2$ gives
$$1 + 4p^6 y^4 - 4y^2 = u^2$$
Let $z = 2p^3$. Since $y \neq 0$ and $z > 2$, we conclude
$$(zy^2 - 1)^2 = z^2 y^4 - 2z y^2 + 1 < z^2 y^4 - 4y^2 + 1 = u^2 < z^2 y^4 = (z y^2)^2.$$
Since $u^2$ lies between two consecutive perfect squares, we get a contradiction.