$\newcommand{Spec}{\operatorname{Spec}}$ Let $R$ be a local Noetherian ring with maximal ideal $m$. Show that if $\Spec(R)\ne\{m\}$ then for every positive integer $n$, $m^n\ne m^{n+1}$. Also what happens with powers of $m$ when $\Spec(R)=\{m\}$?
I would really appreciate some help with this exercise!
The two parts of your question are somewhat related. If $\operatorname{Spec}(R)=\{m\}$, then $R$ is Artinian, and $m$ is its Jacobson radical, so that $m$ is nilpotent and $m^n=0$ for some positive integer $n.$ Thus, the chain of powers of $m$ eventually stabilizes.
As Shivering Soldier suggested, your initial question is easily approached via the contrapositive. If $m^n=m^{n+1}$ for some positive integer $n,$ then Nakayama allows that $m^n=0.$ If $P$ is any prime ideal of $R$, then $m^n \subseteq P.$ But then $m \subseteq P,$ so, since $m$ is maximal, $m=P,$ and $m$ is the unique prime ideal of $R.$ In particular, this tells you that $R$ is Artinian, since all primes are maximal.
All of this is contained in Chapter 8 of Atiyah and Macdonald, at least in my first edition of it.