With GAP it can be verified immediately that there are $303$ groups of order $8892=2^2\cdot 3^2\cdot 13\cdot19$ and also $303$ groups of order $9324=2^2\cdot3^2\cdot 7\cdot 37$.
I do not expect that the number $303$ can be calculated by hand, but perhaps it can be shown by looking at the prime factorizations and using Sylow's theorem, that the number of groups is the same for both orders.
I found out that a group of order $8892$ must have exactly one subgroup of order $13$ and a group of order $9324$ must have exactly one subgroup of order $37$. I do not know whether this helps.
Maybe it is possible to establish a $1-1$ correspondence between a group of order $8892$ and a group of order $9324$, but I have no idea how that can be made.