Yves Daoust's answer covers the two-prime case as well as the three-prime case except when $p \in \{2,3\}$. Here's a simple way to dispense with the remaining cases efficiently:
Suppose $\frac{pqr-1}{(p-1)(q-1)(r-1)}$ is an integer. Then neither $q$ nor $r$ is congruent to $1$ mod $p$.
This is easy to see because the numerator is not divisible by $p$, so the denominator cannot have any factors of $p$.
The above observation immediately eliminates the case $p=2$. For $p=3$, look at the numerator and denominator mod $3$. The numerator is $2\pmod 3$, and so is the denominator (because $q$ and $r$ must be $2$ mod $3$ by above observation). Thus the quotient, if it is an integer, must be $\equiv 1\pmod3$, and since it's $>1$ it must be $\ge 4$. But the computation in Yves's answer shows that it can't possibly be that large.
[Note that $q$ and $r$ are forced to be $2 \pmod 3$ because they can't be $0 \pmod 3$: this is where we use the fact that $q$ and $r$ are prime, so the argument is consistent with the existence of non-prime solutions $(2,4,8)$ and $(3,5,15)$.]