$n=938995200$ also has $22$ solutions. It might be nicer to put $a=y$ and $b=x-y-1$ so $x=a+b+1$ and one has $ab(a+b+1)=n$. For $n=391287046550400$ there are $26$ solutions with $a,b>0$ which grows to $78$ if one allows negative values and shrinks to $13$ if $[a,b]$ and $[b,a]$ are considered the same (and must be positive).
It would seem reasonable that if one chose a number $n$ with "lots" of factors relative to the size of $n$ then any of the curves $ab(a+b+j)=n$ with a "small" j would have a fair number of points and at least some of them would have a large number of points (so one could start with $n$ and look for the most fruitful $j$). That could be made more precise (at least with regard to expectation), maybe not by me though.
later The following sounded plausible but does not turn out to work that well
This suggests seeking $n$ from the highly composite numbers (more divisors than any smaller number.) In fact $391287046550400$ is on that list! (Although I did not know that when I found it) However I tried $n=106858629141264000$ from further down the longer list linked there and only found two points for $j=1$. I did not look at other $j$.
Continued I found $391287046550400$ by looking for products $ab(a+b+1)=n$ with all prime factors under 30 (and no prime over 7 repeated in $n$), and looking for $n$ which turned up frequently. Then I decided to look at the hcn and found that $n$ on the list. However it appears that up to about $1.7 \, 10^{28}$(which is something like the first 260 such ) the appropriate curve has 26 positive points in that one case, 14 in another, and 12 and 10 just a handful of times.
Among those $n$ values, the curve $ab(a+b-7)=481880599200$ has $28$ positive integer points and the curve $ab(a+b+9)=195643523275200$ has $48$ positive integer points but those are the only ones $ab(a+b+j)=n$ which better $ab(a+b+1)=391287046550400$ with a smaller $n$ and $|j|<50$
even later THEN UPDATED Consider these two four integers
$2888071057872000=2^7*3^3*5^3*7*11*13*17*19*23*29*31$
$\ $\begin{eqnarray}
2888071057872000=&&2^7\cdot 3^3\cdot 5^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot 31\\
8659883232000 =&&2^8\cdot 3^3\cdot 5^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 31\\
32607253879200=&&2^{5}\cdot3^{3}\cdot\cdot5^{2}\cdot7^{2}\cdot11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\\
1248124550400=&&2^{8}\cdot 3^{3}\cdot 5^{2}\cdot 7^{2}\cdot 13\cdot 17\cdot 23\cdot 29
\8659883232000 = 2^8*3^3*5^3*7*11*13*17*19*31$end{eqnarray}$$
The first has $32,768$ factors which makes it a hcn since every smaller integer has fewer. The second is the excellent value of $n$ found by tapio which makes $ab(a+b+1)=n$ have 28 positive solutions. The third is also a hcn and the fourth makes $ab(a+b-1)=n$ have 28 positive solutions (or if you prefer, $xy*(x-y+1)=n$ an equation which seems as hard as the chosen one). That is where I would look for similar examples, n a hcn maybe modified by putting in or taking out a couple of large primes and fiddling with the exponent of the smallest primes. Brute calculations will not prove anything of course.
The integer points of $ab(a+b+1)=n$ all transfer to integer points of $ab(a+b+3)=n*3^3$ and that later curve might have additional points, SO maybe it is most sporting to stick to $j=1$.
