= Solution
First establish a stronger form of the <coset>-counting argument. If $K$ is a finitely generated infinite-index <subgroup> of a <finitely generated group> $L$ of growth at most $q$, choose a finite <generating set> $T$ for $K$, and let $M$ bound its lengths in a <generating set> $S$ of $L$. The <Schreier graph> supplies $r+1$ distinct left-<coset> representatives of $K$ of $S$-length at most $r$. The corresponding translates of $B_T(r)$ are disjoint and lie in $B_S((M+1)r)$. Hence
$$
(r+1)|B_T(r)|\le C(1+(M+1)r)^q.
$$
Thus \b[an infinite-index finitely generated <subgroup> has growth at most $q-1$] whenever the ambient upper bound is $q$.
Every finitely generated $K\le H$ has infinite index in $G$. Applying this result with $L=G$ gives growth at most $n-1$. Minimality of $n$ among the supposed counterexamples therefore makes each such $K$ virtually nilpotent. Here $n\ge1$, since a <group> with a uniform bound on its finite word balls is finite, and finite groups are virtually nilpotent.
For each finitely generated $K\le H$, let $d(K)$ be the least nonnegative integer giving a polynomial growth upper bound. It exists and belongs to $\{0,\ldots,n-1\}$. Choose $K$ for which this integer is maximal. If $K\le L\le H$ and $L$ is finitely generated, then $d(L)\le d(K)$. Were $[L:K]$ infinite, the strengthened bound would make $K$ have growth at most $d(L)-1\le d(K)-1$, contradicting the definition of $d(K)$. If $d(K)=0$, the same inequality would force the nonempty balls of $K$ to have cardinality tending to zero, also impossible. Hence every such overgroup has $[L:K]<\infty$.
Since $K$ is virtually nilpotent, choose a nilpotent finite-index <subgroup> $N\le K$. It is finitely generated by <Schreier's lemma>. For any $h\in H$, the <subgroup> $L=\langle K,h\rangle$ is finitely generated and $[L:K]<\infty$. Pigeonholing the cosets $h^jK$ gives a positive $a$ with $h^a\in K$. Pigeonholing the cosets of $N$ in $K$ then gives a positive $b$ with $(h^a)^b\in N$. Consequently
$$
\boxed{\text{For every }h\in H\text{ there is }r=ab>0\text{ with }h^r\in N.}
$$
No <normality> of $K$ in $L$ or of $N$ in $K$ is required. Using least integer upper bounds also avoids needing any further structural theorem about the exact growth degree of a <virtually nilpotent group>.
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