Then 43xy must be a multiple of x^2-xy+y^2.  In other words, 43xy/(x^2-xy+y^2) is an integer.

Let f be the greatest common factor of x and y, then x=fw and y=fz where w and z are coprime.

Then substitute and simplify to get 43wz/(w^2-wz+z^2).  The denominator is coprime to w and z.  Then we must have 43/(w^2-wz+z^2) is an integer.

The only factors of 43 are 1 and 43, so then w^2-wz+z^2=1 or w^2-wz+z^2=43

If w^2-wz+z^2=1 then (w-z)^2 + wz = 1, which can only happen for w=z=1

Then x=y=f reduces the original equation to 2f^3 = 44f^2.  This has one positive root, f=22.  Then x=y=22 is a solution.

If w^2-wz+z^2=43, I note that if both w and z are 7 or larger then the left side is always greater than 43.  So the smaller of w or z must be 1 to 6.  Without loss of generality let z be the smaller.

So substituting z=1 through z=6 and solving finds only one more option, w=7 and z=1.

Then x=7f and z=f reduces the original equation to 344f^3 = 344f^2.  This has one positive root f=1.  Then x=7 and y=1 is a solution.  By symmetry x=1 and y=7 is also a solution.

There is a total of three pairs (x,y) of positive integers that satisfy the given equation: (22,22), (7,1), and (1,7).