I came across the above proposition on mathstackexchange If every nonzero element of $R$ is either a unit or a zero divisor then $R$ contains only finitely many ideals. the link asks a different question but assumes one i ask here. i tried proving the same but i can't. Some hint would be useful. I know basic ring theory. In particular i do not know artinian local rings as someone suggested that it follows easily from the theory of artinian rings.
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Hint: Given $a\in R$, consider the chain of ideals $a^n R$.
Another hint: Given $a\in R$, $a$ not a zero divisor, consider the map $I\mapsto aI$ on the set of ideals of $R$.
Let me help you see the connection to Artinian rings. A commutative ring with finitely many ideals is Artinian, so if you prove it for Artinian rings, then you've proven it for rings with finitely many ideals. Before you, people have asked this question for finite rings rather than for rings with finitely many ideals. Some of the solutions will work in your case.
The approach is to examine the descending chain of ideals of the form $x^iR$ for $i\in \Bbb Z^+$, as in this solution. The Artinian condition (or the fact that there are only finitely many ideals, if you want to stick with it) means that this sequence eventually stops, and then you can show that either $x$ is a zero divisor, or $x$ is a unit.
In fact, it's true that every element of a right Artinian ring is a unit or a zero divisor, but the commutative case is considerably easier. It may be a little advanced for you at this point, but if you are interested in digressing someday, I talked about even more general rings with this property at this related question.