Note that $nk = s^2$ for some $s$. By Factor Lemma, pick $n = ab$, $k = cd$, and $s = ac = bd$. Now we can tile with $a \times b$ rectangles!

Since $nk=m^2$ it is possible to find coprime $u,v$ and $d$ with $n=du^2, k=dv^2, m=duv$. Then consider $u \times du$ tiles. Place $dv$ of them horizontally and $v$ such layers to obtain the square again.

Assume that $n\cdot k = m^{2} $, we have $n | m^{2}$. So there exists a,b such that $n=ab,a | m,b | m$. So we can cut the square into $a\times b$ rectangles.