Getting Back to 1936

we have two cases arising case 1:(604)^2- 9!=1936

case 2:(-604)^2- 9!=1936

we will have maximum value of y- x in case 2 which is 9-(-604)=9+604=613

From the graph we see that max y is 13. y is factorial so $0\leq y \leq 13$ .
Find integer x for a trial integer y. We see from the graph only (x,y)=(604,9), is such a point.
Since we have x as a square, (-604,9) too must be the solution. And max y-x would be with x=-604.
We may use a calculator and evaluate the expression $x^2-y!-1936$ by trial and error.
When both x and y on the curve are integers, the expression will be =0.
If assumed y is not a solution, certain CONSECUTIVE values of x will have opposite signs for the expression.
Assume y=8; For x=203, exp= -1047,.......x=205, exp= -231,.......x=206, exp=+180.
Clearly exp=0 is between x=205 and x=206. So y=8 has no solution.
Assume y=9; For x=603, exp= -1207,.......x=605, exp= +1209,.......x=604, exp=0. It is the one of the solution.