Using $p=p_1$ and $q=p_2$ for the two primes, multiplication by $p^2q^2$ brings it to
Since you want $x=pq...$ a product of distinct primes including $p,q$, write $x=pqy$ where $y$ is to be squarefree and not divisible by either of $p,q$. This brings it to
From this, since neither of $p,q$ divide $y$, we see that $p|a$ and $q|b$. Now put $a=pa'$ and $b=qb'$, and it becomes
From this we see that $pq|c$ and may put $c=pqc'$ and arrive at
Now for solutions, it all depends on what $a',b',c'$ happen to be at this point. Certainly $c'$ would have to be divisible by $y$,and so we could write $c'=yc''$ and then get to
At this point it seems the equation only boils down to the initial values of $a,b,c$, and may end up with no solutions, or with any number of them, by tracing backward through the substitutions above.