Since the page I referred to contains a wrong statement. Here is a complete answer.

Any variety where all finitely generated groups are residually nilpotent has all finitely generated groups residually finite. Now apply Theorem 3.24 of the paper you cite and note that a variety $V$ where all finitely generated groups are residually nilpotent cannot contain $C_p\wr C$. Hence by part 1 of the theorem all finitely generated groups in the variety are virtually nilpotent, hence nilpotent

Here is a more or less detailed (but far from optimal) proof of the last "hence". Suppose that the variety $V$ contains a non-nilpotent finitely generated group $G$. By the assumption, $G$ has a normal nilpotent subgroup $N$ with $G/N$ finite and nilpotent. Hence $G$ is solvable. By Gruenberg's theorem (Th. 2.2 here) $G$ is not Engel. Hence there exists $n\in N$ and $g\in G$ such that $[n,g,g,g,...]$ is never 1. We can assume that the nilpotency class of $N$ is the smallest possible. Then if $N$ is not Abelian, both groups $N$ and $G/[N,N]$ are nilpotent, and by the well known by Theorem 7 of P. Hall , $G$ is nilpotent. Hence we can assume that $N$ is Abelian.

Hence $G$ is Abelian-by-cyclic (generated by the image of $g$). Let $N_0$ be the maximal finite subgroup of $N$. If we can find our element $n$ in $N_0$, we can assume that $N=N_0$, and the group $G$ is finite, hence $G$ is nilpotent. So we can assume that we cannot find such an $n$ in $N_0$. Therefore we can assume that $N_0=\{1\}$ (take $G/N_0$). So $N$ is torsion-free. If some power $g^m$ of $g$ is in $N\setminus \{1\}$, we have that $g^m$ is central, and we can take the quotient by $\langle g^m\rangle$. So we can assume that $C=\langle g\rangle$ intersects $N$ trivially. But that means $g$ is of finite order (since $N$ is of finite index in $G$) and $G$ is a semidirect product of a finitely generated torsion-free Abelian group $N$ by $C$.

The element $g$ acts (by conjugation) on $N$ as a square matrix $A$ of finite size (the rank of $N$) with integer coefficients. Viewing $n$ as an integer vector we get that $[n,g]=An-n=(A-1)n$. Hence $[n,g,g,...] (t\ \mathrm{ times})= (A-1)^tn$. (In particular, the matrix $A-1$ is not nilpotent). Then (by standard linear algebra plus a little number theory) there exists a prime $p\gg 1$ such that $(A-1)^tn$ is never zero modulo $p$. Now let $N^p$ be the $p$-th power of $N$. Then $G/N^p$ is finite, belongs to $V$ and not nilpotent, a contradiction.

**Update.** Here is a little number theory needed in the proof (it corrects the not quite correct comment by @YCor below). First note that it is enough to prove that $A-1 \mod p$ is not nilpotent for some prime $p$ (indeed, then one can take one of the finite set of generators of $N$ for $u$).

Thus we need $p$ such that $trace(\lambda - (A-1)^s)$ modulo $p$ is not equal to $(\lambda-1)^k$ for every integer $s\ge 1$ where $k$ is the size of the matrix $A$ (the rank of $N$). Note that if $(A-1)^m=0 \mod p$ then $(A-1)^{m'}=0$ for every $m'>m$. Now take prime $p$ bigger than the absolute value of every coefficient of the characteristic polynomial of $A-1$. It is enough to show that $(A-1)^{p^m}$ is never $0$ mod $p$. Let $q_s(\lambda)$ be the trace polynomial $\det(\lambda-(A-1)^{p^s})$. Then modulo $p$, $q_s(\lambda^{p^k})=\det(\lambda^{p^s}-(A-1)^{p^s})=\det(\lambda - (A-1))^{p^s}=
q_1(\lambda)^{p^s}=q_1(\lambda^{p^s})$ by the Newton binomial formula and the well known properties of binomial coefficients mod $p$. Since the coefficients of $q_1(\lambda^{p^s})$ are smaller than $p$, this polynomial modulo $p$ is not equal to $(\lambda^{p^{s}}-1)^k$. So $(A-1)^{p^s}$ is not equal to $0$ modulo $p$.

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