Question 570274
<pre>
      xy = 2x - y
  xy + y = 2x
y(x + 1) = 2x
       y = {{{(2x)/(x+1)}}}

The graph of that has vertical asymptote x=-1 and horizontal asymptote y=2.
We plot that graph:

{{{drawing(200,200,-4,4,-2,6, graph(200,200,-4,4,-2,6,(2x)/(x+1) ),
green(line(-10,2,10,2),line(-1,-10,-1,10))
)}}}

The only possibility of y having an integral value when x has an integral
value, is for integral values of x when y is at least 1 unit away from
its horizontal asymptote y=2, and that is when

{{{abs((2x)/(x+1) - 2)}}} {{{"">=""}}} 1

By ordinary methods of college algebra, that has solution 

[-3,-1)U(-1,1]

So we only need to try x-values in that region

which are -3, -2, 0, and 1 

Substituting those in

y = {{{(2x)/(x+1)}}}

we find the only four integral solutions: 

(-3,3),(-2,4),(0,0), (1,1).

So the number of integral solutions is 4.

Edwin</pre>