prime diophantine equation

Find all primes p for which there exists a natural number n such that p^n + 1 is the third power of any natural number.

answer: p=7 is the only solution








proof:
let p^n + 1 = x^3, then p^n = x^3-1 = (x-1)(x^2+x+1)
since this is a prime power, each factor must also be a prime power, so x-1 = p^a and x^2+x+1 = p^b
x^2+x+1 > x-1 for all positive integers x, so we must have p^b > p^a
also p^b/p^a = p^(b-a) must be an integer, so (x^2+x+1)/(x-1) is also an integer
dividing the polynomials, we get p^(b-a) = x + 2 + 3/(x-1) which is an integer
so x+2 is an integer, so 3/(x-1) must also be an integer
so x-1 = 1 or 3, so x=2 or x=4
if x=2, then x^3-1 = 7, so p=7 and n=1
if x=4 then x^3-1 = 63 which is not a prime power
so p=7 is the only solution