Is it possible to draw a cone which never intersects any grid points?

Question

I am wondering if it is possible to define first a line, $z = my + b$ such that when the line is revolved around the $z$ axis to form a cone, this cone never intersects any integer grid points.

In other words, is it possible to define a surface of revolution by the above method such that there are no points $(a,b,c)$, with $a,b,c \in \mathbb{Z}$ which intersect the surface?

What about a cone which only intersects the origin?

There is a simple proof to show that a line $z = m y$, with $m \notin \mathbb{Q}$ [edited from $m \notin \mathbb{Z}$ ] never intersects any integer grid points (other than (0,0)), but I am not sure if it is very clear how to generalise this to a surface of revolution.

Answer 1

Consider the cone $z=\pi +\sqrt {x^2 +y^2}$.

This cone does not have a point with integral coordinates at all. Otherwise $\pi$ would be rational.

Answer 2

$$x^2+y^2=\left(z+\frac12\right)^2$$ has no integer solutions.