I've seen some interesting geometric interpretations of Diophantine equations before. For instance, if we wish to find Pythagorean triples, we can take $a^2 + b^2 = c^2$, divide through by $c^2$, and get $x^2 + y^2 = 1$, which is the unit circle. We then choose an obvious rational point on the circle, e.g., $(-1, 0)$.
Now, any line with a rational finite slope that passes through $(-1,0)$ will intersect the circle at another point with rational coordinates. Then $a$ and $b$ are the numerators of the coordinates, and $c$ is the denominator (after bringing them to a common denominator).
Of course, something similar happens on elliptic curves as well.
I've been looking into a similar problem, but it requires three-variables--and therefore a surface more complex than the unit circle. It's also a cubic:
$$z^3 - z = xy(x^2-y^2)$$
(This is in relation to the order-3 magic square of squares problem.) If a line passes through one rational point on the surface, must any other intersections also be rational points? Or is there a restriction of some sort?
No, the other two points need not be rational. In general they will be defined over a quadratic extension of $\mathbb{Q}$ and conjugate.