Let $R$ be a finite local ring with maximal ideal $M$ such that $|R|/|M|\equiv 1\pmod{4}$. Then $-1$ is a square in $R^\times$ (that is, there exists $u\in R^\times$ such that $u^2=-1$) if and only if $-1$ is a square in $(R/M)^\times$. Is this correct?
I know that it is valid for $\mathbb{Z}_{p^\alpha}$ with $p$ odd prime. Is it correct for other cases?
On
No the completeness of the $p$-adics is important. For example let $\mathbb{Z}_{(p)}$ be the localization of$\mathbb{Z}$ at $p$, ie all $\frac{a}{b}$ with $p\nmid b$.
Then $\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)}= \mathbb{Z}/p\mathbb{Z}$ and $-1$ is a root here but clearly not in $\mathbb{Z}_{(p)}$.
Yes, it's correct!
$R^{\times}\to (R/M)^{\times}$ is a surjective group morphism whose kernel is $1+M$, hence $$R^{\times}/(1+M)\simeq (R/M)^{\times}.$$
If $-1$ is a square in $(R/M)^{\times}$, then there is $a\in R^{\times}$ such that $-a^2\in 1+M$. But $1+M=(1+M)^2$: if $|R|=p^t$, then note that $|1+M|=|M|$, and the last one is also a power of $p$ which is an odd prime. Thus $-a^2\in(1+M)^2$, and therefore there is $b\in 1+M$ such that $-a^2=b^2$, equivalently $-1=(ba^{-1})^2$.
Remark. In fact, $1+M$ is the Sylow $p$-subgroup of $R^{\times}$, and we have $$R^{\times}\simeq (R/M)^{\times}\times(1+M).$$