Let $p,q$ be odd distinct primes. Is it possible to prove with elementary methods that there are no coprime positive integers $a,b$ and an integer $x \ge 2$ such that $$ ab\le 4^p,\,\,\,\,x-1=a^q, \,\,\,\,\text{ and }\,\,\,\frac{x^p-1}{x-1}=b^q? $$
Of course, this is a special instance of Mihailescu's theorem. Probably the first condition $ab\le 4^p$ is not even necessary.
A result with elementary proof which allows easily to conclude (without the need of the first condition) is
J.W.S. Cassels, On the equation $a^x-b^y=1$, II, Proc. Cambridge Philos. Soc. 56 (1960), 97-103
However, this proves much stronger, i.e., that if $x^p-y^q=1$ then $p\mid y$ and $q\mid x$. This implies that $$ p\mid y^q= (x-1)\left(\frac{x^p-1}{x-1}\right), $$ hence at least one of the two factors are divisible by $p$. If $x^p\equiv 1 \pmod{p}$ then also $x\equiv 1\pmod{p}$. Conversely, if $x\equiv 1\pmod{p}$ then $p$ divides $(x^p-1)/(x-1)$ by Lifting the Exponent lemma. Hence we would contradict the coprimality of $a$ and $b$.
Does there exist an easier proof of the result?