Consider the quadratic $n^2-(kn)+m^2-km^4 = 0$ whose discriminant is \[ \Delta = k^2+4km^4 - 4m^2. \]Note that for some $k$ the system has a positive integral solution if and only if there is an $m$ for which the discriminant above is a perfect square. With this, let's now prove that if $k=p^2$ where $p\equiv 3\pmod{4}$ is a prime, then the system has no solutions. Suppose the contrary. Then for $k=p^2$, $p\equiv 3\pmod{4}$ and $\ell\in\mathbb{N}$, \[ p^4 + 4p^2 m^4 - 4m^2 = \ell^2. \]Using modulo $p$, we get $\ell^2\equiv -(2m)^2\pmod{p}$. Since $(-1/p)=1$, we must have $p\mid m$ and $p\mid \ell$. Set $m=p\alpha$ and $\ell = p \ell_1$ to arrive at \[ p^4 + 4p^2 \cdot p^4 \alpha^4 -4p^2\alpha^2 = p^2 \ell_1^2 \Rightarrow p^2+4p^4 \alpha^4-4\alpha^2 = \ell_1^2. \]Once again, using modulo $p$, we find $p\mid \alpha,\ell_1$. Set $\alpha=p\beta$ and $\ell_1=p\ell_2$. We then arrive at \[ p^2 + 4p^4 \cdot p^4 \beta^4 - 4p^2 \beta^2 = p^2 \ell_2^2 \Rightarrow 4p^6\beta^4 -4\beta^2 + 1 =\ell^2. \]Now, observe that \[ \left(2p^3 \beta^2-1\right)^2 < 4p^6 \beta^4 - 4\beta^2+1 < \left(2p^3\beta^2\right)^2, \]so the given expression cannot be a perfect square. Credit. Thanks to @electrovector for suggesting to work `backwards'.

There was a nt kind inequality on previous year, expected to see one