arithmetic geometry - Integer points (very naive question)

To make sense of the notion of integer points, your scheme should be defined over $mathbb{Z}$. What do we mean by that? Of course we should not ask for a structure map tp $Spec(mathbb{Z})$, since every scheme has one such map. The right notion is the following.

Let $X$ be a scheme over $mathbb{C}$; so by definition we have a structure map $X to mathop{Spec} mathbb{C}$. Then we say that $X$ is defined over $mathbb{Z}$ is there exists a scheme $X_{mathbb{Z}}$ over $mathbb{Z}$ such that $X$ is the base change of $X_{mathbb{Z}}$ to $mathbb{C}$, i. e. $X cong X_{mathbb{Z}} times_{mathbb{Z}} mathop{Spec} mathbb{C}$.

Now for such a scheme an integral point is a map $mathop{Spec}mathbb{Z} to X_{mathbb{Z}}$ such that the composition with the structure map is the identity. Note that the same can be done for every ring $A$ in place of $mathbb{Z}$.

With this definition, the line ${ x = 0 }$ is defined over $mathbb{Z}$, but the line ${ x = pi }$ is not, basically because there is no way to generate its ideal with equations having integer coefficients. So your problem does not arise anymore.

EDIT: Abstractly of course the two lines are isomorphic over $mathbb{C}$, so the line $r = {x = pi }$ actually has a model over $mathbb{Z}$. The problem is that this model is not compatible with the inclusion in $mathbb{A}^2$, that is, there will be no map $r_{mathbb{Z}} to mathbb{A}^2_mathbb{Z}$ whose base change is the inclusion of $r$ into $mathbb{A}^2$. In order to have this, you would have to ask that the ideal of $r$ in $mathbb{A}^2$ should be generated by polynomials with integer coefficients.

As for your second question, there can be different models, that is, nonisomorphic schemes over $mathbb{Z}$ which become isomorphic after base change to $mathbb{C}$. So before discussing the existence of integral points, you have to FIX a model, and the points will in general depend on the model.

For instance take the two conics ${ x^2 + y^2 = 2 }$ and ${ x^2 + y^2 = 3 }$. Both have an obvious choice of a model, given by the inclusion in $mathbb{A}^2$; moreover they are isomorphic over $mathbb{C}$. But the integral points on the first one are $(pm 1, pm 1)$, while the second has none.

Finally you consider the possibility that the structure over $mathbb{C}$ is not relevant. This is false: the base change $X_mathbb{Z} times_{mathbb{Z}} mathop{Spec} mathbb{C}$ is endowed with a natural map to $mathop{Spec} mathbb{C}$, and we ask for the isomorphism with $X$ to be over $mathbb{C}$.