Is there a field k such that $Spec(k)\times \mathbb P^1$ is affine?
Probably I have no technical background to claim any good idea on this. However, I think the answer should be negative since $Spec(k)=\{(0)\}$ and topologically thinking $Spec(k)\times \mathbb P^1$ homeomorphic to $\mathbb P^1$ and we know that the scheme $\mathbb P^1$ is not affine.
How to correctly show this, and is my explanation so stupid?
Note that over an infinite field $k$, $\mathbb{P}^1_k$ is homeomorphic to $\mathbb{A}^1_k=\text{Spec } k[x]$. Being affine is not a topological property.
The best way, in my opinion, to see that the projective line is not affine is to compute the global regular functions. There are no non-constant global functions over $\mathbb{P}^n_k$ (see for example, this question), so if it were affine, it would necessarily be equal to $\text{Spec } k$.