Rewritten the equation and solve $y$ with respect to $x$: $$y=\frac{x^2 + 2x}{x+3}=\frac{x^2 +3x - x}{x+3}=x -\frac{x+3-3}{x+3}=x-1 + \frac{3}{x+3}$$

Because $y$ is an integer, $x + 3=\pm 1, \pm 3\implies x=-4, -2, -6, 0$. Accordingly, $$(x, y) = (-6, -8),(-4, -8),(-2,0),(0, 0)$$