Solve in $\mathbb{Z}$, then in $\mathbb{Q}$, the equation: $2 x^{3}+x y-7=0$
The equation can be written as: $x\left(2 x^{2}+y\right)=7$. In $\mathbb{Z}$, this requires that $x$ be a divisor of 7, in which case $y=\frac{7}{x}-2 x^{2}$, leading to four possibilities:

$x=1, y=5, x=-1, y=-9, x=7, y=-97, x=-7, y=-99$.

In $\mathbb{Q}$, $x$ can be any (non-zero) value: for any non-zero rational $x$, $y=\frac{7}{x}-2 x^{2}$ is a rational number such that $(x, y)$ satisfies the equation.