= Solution
For a <group presentation> $\mathcal P=\langle X\mid R\rangle$ with $X$ finite, let $F(X)$ be its <free group>. For a nontrivial relator define
$$
\nu_p(r)=\max\{k\geq0:r=w^{p^k}\text{ for some }w\in F(X)\}.
$$
The <p-deficiency> in the unshifted convention used here is
$$
\boxed{\operatorname{def}_p(\mathcal P)=|X|-\sum_{r\in R}p^{-\nu_p(r)}.}
$$
If the weighted sum diverges the value is $-\infty$; identity relators may be omitted or assigned weight zero. Roots are taken in the <free group>, not in the presented quotient. Some authors subtract one from this definition; here the requested threshold is $1$.
Two elementary bounds explain why <p-deficiency> detects infinitude. The <p-rank of a group> is
$$
d_p(G)=\dim_{\mathbb F_p}\bigl(G/[G,G]G^p\bigr).
$$
Here $G^p$ denotes the subgroup generated by all $p$th powers. Each relator that is not a $p$th power imposes at most one linear relation in this <vector space>, and a $p$th-power relator imposes none. If there are $s$ relators of the first type, then
$$
d_p(G)\geq |X|-s\geq\operatorname{def}_p(\mathcal P).
$$
The second bound is the <index-p rewriting bound for p-deficiency>. Suppose $H\trianglelefteq G$ has index $p$, and its preimage in $F(X)$ is $V$. The <Nielsen–Schreier formula> gives $V$ rank $1+p(|X|-1)$. For a relator $r=w^{p^k}$, there are two cases in the <Reidemeister–Schreier theorem>. If $w\in V$, its $p$ coset-conjugates are all $p^k$th powers in $V$, with total weight at most $p\,p^{-k}$. If $w\notin V$, then $k\geq1$, since $r\in V$. Its cosets generate $F(X)/V$, so representatives $1,w,\ldots,w^{p-1}$ show that the $p$ rewritten conjugates of $r$ are redundant up to conjugation in $V$. One relator suffices, and $r=(w^p)^{p^{k-1}}$ has weight at most $p^{-(k-1)}$. In both cases the total weight is at most $p$ times the old weight. Thus the induced <group presentation> $\mathcal Q$ of $H$ satisfies
$$
\boxed{\operatorname{def}_p(\mathcal Q)-1\geq p\bigl(\operatorname{def}_p(\mathcal P)-1\bigr).}
$$
The argument applies termwise to infinitely many relators whenever the weighted sum converges.
If $\operatorname{def}_p(\mathcal P)\geq1$, the <p-rank of a group> bound gives a surjection to $C_p$, hence a normal subgroup of index $p$. The rewriting bound gives that subgroup another presentation of <p-deficiency> at least one. Iterating produces subgroups of index $p^j$ for every $j$. \b[<p-deficiency at least one implies infinitude>.]
Now enumerate the nonidentity elements $w_1,w_2,\ldots$ of $F(x,y)$ and choose the presentation
$$
T=\langle x,y\mid w_i^{p^{i+2}}=1\ (i\geq1)\rangle.
$$
Its <p-deficiency> obeys
$$
\operatorname{def}_p(T)\geq2-\sum_{i\geq1}p^{-(i+2)}=2-\frac1{p^2(p-1)}>1.
$$
The infinitude criterion shows that $T$ is infinite. It is generated by two elements, and every element is represented by some $w_i$ or is the identity; the imposed relation makes its order a power of $p$. Thus \b[$\boxed{T\text{ is an infinite finitely generated torsion group}}$]. This is a <torsion group construction by p-power relators>; the presentation intentionally has infinitely many relators.
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