Let $A$ be a nonzero commutative ring with unit. Define $\mathbb P_A^n$ to be the scheme $\operatorname {Proj} A[T_0,\dots,T_n]$, where the grading on the polynomial ring is by degree. Why is it true that $\mathbb P_A^n\not\cong \mathbb P_A^m$ when $n\neq m$? (I admit I am just guessing here. It seems like it should be true. But I do not know why.)
I am still learning the basics of scheme theory and would prefer as simple an answer as possible. Can this be proved directly from the definitions, without much machinery?
My motivation for asking this question is to show that the only $n$ for which $\mathbb P_A^n$ is affine is $n=0$. I can show that for any $\mathbb P_A^n$, the sheaf of global sections is $A$. So it is easy to see that if $\mathbb P_A^n$ is affine, the only possibility is that it is isomorphic to $\operatorname{Spec} A$. One can directly verify that they are isomorphic for $n=0$, but I do not know how to rule out larger $n$. Hence this question.
There is a certain sense in which the dimension tells them apart. This can be made precise by a cohomological argument: Thm. 5.1 of Chapter 3 of Hartshorne implies that there is a coherent sheaf on $\mathbb{P}^n$ with a non-zero cohomology group in dimension $n$. Meanwhile, if $m<n$, then $\mathbb{P}^m$ has a covering by $m+1$ affine schemes, all of whose intersections are also affine, and this implies that no coherent sheaf on $\mathbb{P}^m$ has cohomology in any dimension bigger than $m$ (start with Thm 4.5 of Hartshorne).