All we really need are the largest power of 2 and the largest power of 3 which divide 19^88-1

First factor 19^88-1 = (19^44+1)*(19^22+1)*(19^11+1)*(19^11-1).

Lets take this mod 8.  19^11 = 3^11 = 3*9^11 = 3*1^11 = 3 mod 8.  Then substituting this the four factors are congruent to 3^4+1=2, 3^2+1=2, 3+1=4, and 3-1=2 mod 8.  Then the largest power of 2 which divides 19^88-1 is 2^5

Now we want to consider this factorization mod 9.  19^11 = 1^11 = 1 mod 9.  So them the only factor which is a multiple of 3 is 19^11-1.

Factor this further into (19-1)*(19^10+19^9+...+19+1).  The large factor is congruent to 2 mod 9, so then 19-1=18=2*3^2 leaves us with the largest power of 3 which divides 19^88-1 is 3^2.

The sum of all the factors of the form 2^a*3^b with a>0 and b>0 can be expressed as (2^5+2^4+2^3+2^2+2)*(3^2+3) = 744.