By Meyer's theorem, an indefinite quadratic form $Q$ over $Z$ has an integral isotropic vector is the dimension is at least $5$ and this bound is tight. Indeed, in dimension $4$ there are indefinite quadratic forms without isotropic vectors.
What happens if we replace $Z$ by a ring of integers? One interesting example could be $R = Z[\sqrt{2}]$. I expect the following to be true:
There exists an integer $n_0$ such that:
For any indefinite quadratic form $Q$ over $R$ of dimension $n\geq n_0$ there exists an isotropic vector over $R$.
For any $n<n_0$, there exists an indefinite quadratic form $Q$ of dimension $r$ over $R$ such that $Q$ has no isotropic vector over $R$.
If true what would be the dimension $n_0$? How the answer would vary over a real algebraic number ring $R$?
