= Solution
We construct an <infinite residually finite torsion group> using the <Golod-Shafarevich infinitude criterion for pro-p groups>. Its useful form says that a pro-$p$ presentation on $d$ generators with relators $r_i$ gives an infinite <pro-p group> whenever
$$
1-d\tau+\sum_i\tau^{D(r_i)}<0\qquad\text{for some }0<\tau<1.
$$
Here $D$ is the <Magnus degree of a free pro-p group element>: map each generator to $1+X_j$ in a noncommutative power-series algebra over $\mathbb F_p$, and take the lowest degree of $w-1$. The criterion permits infinitely many relators with a convergent weight sum. It is a consequence of the Golod-Shafarevich inequality for the completed group algebra; see https://m-ershov.github.io/Research/gssurvey_revised.pdf[Ershov's survey, Theorem 3.1].
Take $p=2$, $d=2$, and $\tau=3/4$. Let $F$ be the <free pro-p group> on $a,b$, and enumerate all ordinary finite words $w_1,w_2,\ldots$ in these generators and their inverses. Choose integers $e_i$ sufficiently large that
$$
\tau^{2^{e_i}}<2^{-i-2}.
$$
Form the quotient <pro-p group>
$$
P=F/\overline{\langle\!\langle w_i^{2^{e_i}}:i\ge1\rangle\!\rangle},
$$
where the bar means the closed <normal closure>. If the Magnus image of $w_i$ is $1+v_i$, characteristic two gives
$$
(1+v_i)^{2^{e_i}}-1=v_i^{2^{e_i}},\qquad D(w_i^{2^{e_i}})\ge2^{e_i}D(w_i)\ge2^{e_i}
$$
for nonidentity $w_i$; identity relators contribute zero weight. It follows that
$$
1-2\tau+\sum_i\tau^{D(w_i^{2^{e_i}})}\le1-\frac32+\sum_{i\ge1}2^{-i-2}=-\frac14<0.
$$
Thus the <Golod-Shafarevich infinitude criterion for pro-p groups> proves that $P$ is infinite.
Let $G=\langle\bar a,\bar b\rangle$ be the abstract subgroup generated by the two images inside $P$. This subgroup is dense because the two images topologically generate the quotient, and it is infinite: a finite subgroup of a Hausdorff <profinite group> is closed, so could not be dense in infinite $P$. Every element of $G$ is the image of some ordinary word $w_i$, and the imposed relation makes its order divide $2^{e_i}$. Therefore $G$ is a two-generated <torsion group>. Finally $P$ is an <inverse limit> of finite 2-groups. Any nonidentity element of $G$ is nonidentity in $P$ and hence has a nonidentity image in one of those finite groups. Restricting that map to $G$ proves that it is a <residually finite group>. We have obtained
$$
\boxed{G\text{ is finitely generated, infinite, residually finite, and every element has finite order}.}
$$
The abstract subgroup $G$ is essential here: the argument establishes finite order for ordinary words, not for every topological limit in the ambient <pro-p group>.
There is \b[no surjective homomorphism from this group to an infinite soluble group]. In fact <finitely generated soluble torsion groups are finite>. To prove this, let $S$ be finitely generated, soluble and torsion, and induct on its derived length. In the abelian case, finitely many generators of finite orders generate a finite group. In general the <abelianization> $S/S'$ is a finitely generated abelian torsion group, hence finite. Its kernel $S'$ therefore has finite index and is finitely generated by <Schreier's lemma>. Explicitly, for a finite coset transversal $T$ containing $1$ and a finite symmetric generating set $X$, the finitely many elements $tx\overline{tx}^{-1}$ generate the finite-index subgroup, where the bar denotes the chosen representative of the coset. Now $S'$ is torsion and has smaller derived length, so induction makes it finite. Hence $S$ is finite as well. Every quotient of $G$ is finitely generated and torsion, so any soluble quotient must be finite.
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