Mochizuki has already claimed to have proven the ABC-conjecture. But even if his claim turns out to be correct the proof will not be easy to understand. With that in mind I'm asking wether anything is known on the following weaker version of the ABC-conjecture that does not use Mochizuki's work.

Let $S = \{p_1,\ldots, p_k \}$ be a finite set of primes and call an integer $n$ to be $S$-smooth if all prime factors of $n$ are contained in $S$.

My main question now is: Fix a finite set of primes $S$ and an $\varepsilon >0$, then is the number of solutions to $$a+b=c, \quad c > \textrm{rad}(abc)^{1+\varepsilon}$$ with $a$ and $b$ coprime and $S$-smooth integers finite?

Note that since in this case we have $\textrm{rad}(ab)\leq \prod_{p \in S} p$ this question is equivalent to:

Fix a finite set of primes $S$ and an $\varepsilon >0$, then is the number of solutions to $$a+b=c, \quad c > \textrm{rad}(c)^{1+\varepsilon}$$ with $a$ and $b$ coprime and $S$-smooth integers finite?

The baby case where $\#S =1$ is also aready interesting to me. This comes down to the following, fix a prime $p$ and a $\varepsilon >0$. Then is there an $N$ such that for all $n>N$ one has that $p^n \pm 1 < rad(p^n \pm1)^{1+\varepsilon}$?

Note that I'm not requiring $c$ to be $S$-smooth, so it is not a consequence of the $S$-unit equation.