Mehtaab Sawhney has shared the following simple argument that proves that the above limit points are in fact the only ones.

If $v_p(m)$ is the largest $k$ such that $p^k\mid m$ then $\tau(m)=\prod_p (v_p(m)+1)$ and so \[\frac{\tau((n+1)!)}{\tau(n!)} = \prod_{p|n+1}\left(1+\frac{v_p(n+1)}{v_p(n!)+1}\right).\] Note that $v_p(n!)\geq n/p$, and furthermore $n+1$ has $<\log n$ prime divisors, each of which satisfy $v_p(n+1)<\log n$. It follows that the contribution from $p\leq n^{2/3}$ is at most \[\left(1+\frac{\log n}{n^{1/3}}\right)^{\log n}\leq 1+o(1).\]

There is at most one $p\mid n+1$ with $p\geq n^{2/3}$ which (if present) contributes exactly \[\left(1+\frac{1}{\frac{n+1}{p}}\right).\] We have proved the claim, since these two facts combined show that the ratio in question is either $1+o(1)$ or $1+1/k+o(1)$, the latter occurring if $n+1=pk$ for some $p>n^{2/3}$.

After receiving Sawhney's argument I found that this had already been proved, with essentially the same argument, by Erdős, Graham, Ivić, and Pomerance [EGIP].