= Solution
All commutators of $H$ with $K$ lie in the central <subgroup> $Y$ of $HK$. In the standard <commutator> identities, <conjugation> by elements of $H$ or $K$ therefore acts trivially on these commutators. Since $Y$ is abelian, their order can also be reversed. Consequently
$$
[h_1h_2,k]=[h_1,k][h_2,k],\qquad
[h,k_1k_2]=[h,k_1][h,k_2].
$$
Thus the map is a <homomorphism> in each variable separately; this does not assert that it is a <homomorphism> from the direct product jointly.
It factors through $H/C_H(K)$ in the first variable and $K/C_K(H)$ in the second. If these finite <quotient> groups have representatives $h_1,\ldots,h_a$ and $k_1,\ldots,k_b$, then the finite collection $[h_i,k_j]$ generates $Y$. Let $e$ be an exponent of $H/C_H(K)$. For every $h,k$,
$$
[h,k]^e=[h^e,k]=1.
$$
Hence $Y$ is a finitely generated <abelian group> of bounded exponent, and \b[$Y$ is finite]. For instance its order is at most $e^{ab}$.
Now suppose $G$ is nilpotent and $G/\gamma_2(G)$ is finite. Induct on the <nilpotency class> $c$. For $c=1$, $G$ is its own <abelianization> and is finite. If $c\ge2$, the <group> $\bar G=G/\gamma_c(G)$ has smaller class and
$$
\bar G/\gamma_2(\bar G)\cong G/\gamma_2(G),
$$
since $\gamma_c(G)\le\gamma_2(G)$. By induction $\bar G$ is finite. Apply the preceding result with $H=G=\gamma_1(G)$ and $K=\gamma_{c-1}(G)$. Their <commutator subgroup> is $Y=\gamma_c(G)\le Z(G)$. Both relevant <centralizer> quotients are finite, because their centralizers contain $\gamma_c(G)$ and $G/\gamma_c(G)$ is finite. Thus $\gamma_c(G)$ is finite. The extension of this finite <normal subgroup> by the finite <quotient> $\bar G$ is finite. The converse follows because a <quotient> of a <finite group> is finite. Therefore
$$
\boxed{G\text{ finite}\iff G/\gamma_2(G)\text{ finite}\qquad(G\text{ nilpotent}).}
$$
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