= Solution
Within the approximation that a factory has either zero or one accident, let $N$ be the accident count. The factories' independence gives $N\sim\operatorname{Binomial}(m,p)$. Given an accident, its cluster of claims has a <Poisson distribution> of mean $\lambda$, independently across factories. Its <probability generating function> is $G_X(z)=e^{\lambda(z-1)}$. Thus this is a <binomial accident portfolio with Poisson clusters>, with
$$
G_S(z)=\left(q+pe^{\lambda(z-1)}\right)^m,\qquad\boxed{\mathbb P(S=0)=\left(q+pe^{-\lambda}\right)^m.}
$$
A factory can contribute no claims either because it has no accident or because its accident's Poisson cluster is empty. Hence the no-claim probability differs from the no-accident probability $q^m$.
For $0\leq n\leq m$, the accident <probability mass function> is $p_n=\binom mn p^nq^{m-n}$. For $1\leq n\leq m$,
$$
\frac{p_n}{p_{n-1}}=\frac{m-n+1}{n}\frac pq=-\frac pq+\frac{(m+1)p}{nq}.
$$
Therefore its <Panjer claim-count class> parameters are
$$
\boxed{a=-\frac pq,\qquad b=\frac{(m+1)p}{q}.}
$$
At $n=m+1$ the multiplier is zero, and afterwards both successive masses are zero, so the recurrence holds for all $n\geq1$. Finally, $f_0=e^{-\lambda}$ and $G_N(z)=(q+pz)^m$ reproduce the same initialization $g_0=G_N(f_0)$.
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