Let $p$ be an odd prime number.
Ramanujan's tau function satisfies:


(a)
$$
\tau(p^{n+1}) = \tau(p^n)\tau(p)-p^{11}\tau(p^{n-1})
$$
for all positive integers $n>0.$
So $\tau(p)=0$ implies


(b)
$$
\tau(p^{2r+1})=0,
$$
and
(c)
$$
\tau(p^{2r})=(-1)^rp^{11r}
$$
for all nonnegative integers $r \geq 0.$


Assume now that (b) happens for {\it{some}} $r \geq 0.$


Question: Can we get $\tau(p)=0$ ?


We may assume from classic Lehmer's result that $n=p^{2r+1}$ is {\it{not}} the smallest $n$ with $\tau(n)=0.$


Seems that adding condition (c) for the same $r$ works, since (essentially): if $p^k || \tau(p)$
then $k$ should be very small.



If Ramanujan's tau function has a prime power zero then $\ldots$
Theorem 2 of Lehmer's paper "The vanishing of Ramanujan's function $\tau(n)$" says that the smallest $n$ for which $\tau(n)=0$ is prime. 


More generally, this paper should be of interest to you given your question.


