This is a very nice problem. My first idea was to see if this problem had no solution and prove it with mod but that didn't work out very nicely. So, I solved for $y$ in terms of $x$. After moving things around, I got $y = \frac{6x^2-13x+11}{3x-5}$. Using Long Division, this becomes $2x-1 + \frac{6}{3x-5}$. For $y$ to be an integer, we need to have $3x-5$ as an integer divisior of 6. So, $3x-5 = \pm 1, \pm 2,\pm 3,\pm 6$. Solving for $x$, we only get 2 integer $x$ values and plugging them gives $y$ values as well. Those are $(2,9), (1,-2) \blacksquare$.