Here is a simpler bound, based on the comment of R. van Dobben de Bruyn.
Let a solution of the equation be broken into two parts, c and d, where c is the n-smooth part of the solution, and is coprime to d, which of necessity is square free and has all prime factors bigger than n. (I leave the case n=1 to the reader.) Then c is at most n! (2/1)(3/2)(5/4)...((n+1)/n)which for n greater than three is strictly less than (n+1)n!. So c is less than (n+1)! .
Turning to d, each prime divisor of d contributes at least one power of 2 when subject to Euler's phi, so d has fewer than n prime divisors. So d is less than e times n!.
Since the shrinkage under phi of the product is at most e(n+1), the original solution must be less than 3(n+1)!. This is also a weak upper bound on the total number of solutions, but can probably be improved to show that the number and location of solutions generally is less than (n+1)!, leaving the case of small n to the reader where all creation (counterexamples, arrghh spellcheck!) must lie.
Gerhard "Leaving Hard Work To Others" Paseman, 2020.07.07.