The first equation implies $x + y^2 = xy + 1$, or $x - xy = 1 - y^2$, so $x(1 - y) = 1 - y^2$. This can be rewritten as $(1 - y)(x - 1 - y) = 0$.

If $y = 1$, then this equation is satisfied. The second equation then implies $x \cdot 1 = 4 + 1$, so $x = 5$.

If $y \ne 1$, then $x - 1 - y = 0$, so $x = 1 + y$. Combining this with the second equation yields:

$(1 + y)y = 4 + y$

$y^2 + y = 4 + y$

$y^2 = 4$

So, $y = 2$ or $y = -2$.

For $y = 2$, $x = 1 + 2 = 3$.
For $y = -2$, $x = 1 + (-2) = -1$.

Both equations are satisfied for $x = 5$, $y = 1$; for $x = 3$, $y = 2$; and for $x = -1$, $y = -2$.