I heard a great answer to this question based on the Thue equation.

I investigated the Thue equation and there was one point that was not clear to me. It seems to me that there are an infinite number of values that $a$ and $b$ can take. If there are an infinite number of combinations of finite solutions, then there is an infinite number of solutions. Right?

So, if I understand it, the Thue equation alone doesn't seem to work. I apologize if I am misunderstanding the classical result there.

Here's an argument that seems to work as far as I understand:

(1) For any prime $p$, there is a finite number of combinations of primes that are less than $p$.

(2) For any of these combinations there exists an integer $x$ such that if a combination is greater than $x$, then at least one of the primes that make up the combination are of a degree greater than $2$.

(3) Let $c$ be either the highest of the values $x$ from step (2) or $4p^{4}(3\prod_{p} p^{\frac{1}{2}})^3$ depending on which is higher.

(4) There is a finite number of ways that we can pair these different combinations so that we have an equation of the form $ax^3 - by^3 = c$ where $c < p$.

(5) If $a \ne b$ and $gcd(x,y)=1$, then using a result from Siegel, there is at most $1$ solution (see Theorem A in the link below).

http://matwbn.icm.edu.pl/ksiazki/aa/aa75/aa7538.pdf

(6) if $a = b$ and $a = c$, then using a result from Michael Bennett, there is at most $1$ solution. Here's the reference:

M. A. Bennett, Rational approximation to algebraic numbers of small height : the Diophantine equation $|ax^n-by^n|=1$}, J. Reine Angew. Math. 535 (2001), 1-49

http://www.math.ubc.ca/~bennett/B-Crelle2.pdf

(7) if $a = b$ and $a < c$, then the equation has a form such as:

$x^3 - y^3 = \frac{c}{a}$

This is a Thue Equation and we can conclude that there is a finite number of solutions.

(8) if $a \ne b$ and $gcd(x,y) > 1$, then it means that both combinations consist of the same prime so we have an equation of the form:

$x^m - x^n = c$

Then, as I understand it, we have a Thue equation so we can again assume that there is a finite number solutions.

I believe that covers all the possible cases.

Since, there are only a finite number of solutions, it follows that there exists an integer $c$ which is greater than all of these solutions and for all $x \ge c$, we have at most $1$ integer in the sequence where $gpc(x+i) \le p$.

Apologies for the length of this argument. I'm sure a professional mathematician such as user 631 would be able to state the argument more elegantly. :-)