Let $R$ be a finite integral domain, and $a$ be a nonzero element of $R$. Why is it true that there are distinct $m, n$ such that $a^m = a^n$ in $R$? Let $R$ be a finite integral domain, and $a$ be a nonzero element of $R$. Why is it true that there are distinct $m, n$ such that $a^m = a^n$ in $R$?
Does the fact that $R$ is finite imply that it must also be cyclic? If it does, then I understand the answer to the question I asked, but then why must it be cyclic? Must all finite groups be cyclic? All finite rings?
I am quite new at this, and I am feeling pretty confused. Any help would be appreciated.
 A: If there are no such $m$ and $n$, then the elements $a,a^2,a^3,a^4,\dots$ are all distinct elements of $R$.  But these elements then form an infinite subset of $R$, which contradicts the assumption that $R$ was finite.  (Note that this does not require $R$ to be a domain or $a$ to be nonzero.)
A: 
Must all finite groups be cyclic? 

Certainly not. Just look at the Klein 4-group. To see a little more clearly why, just remember that a cyclic group of order $n$ must contain an element of order $n$, but as you see in the case of the Klein 4-group, all nonidentity elements have order $2$, and none have order $4$.

All finite rings? (=Does the fact that $R$ is finite imply that it must also be cyclic?)

There is a double bit of confusion here: what does it mean for a ring to be cyclic? Apparently you are using it to mean that it is multiplicatively generated by a single element. But this can't ever be the case for a ring with nonzero identity: there must be minimal $n$ and a minimal $m$ such that $a^n=1$ and $a^m=0$. If $m<n$ then $0=1$, and if $n<m$ you have $a^m=1\cdot a^{m-n}=0$, but $m-n$ is less than m, contradicting minimality of m. Another way to look at it is that $a^n=1$ implies a is a unit, but $a^m=0$ implies it s a zero divisor, which is contradictory.
On the other hand, it is possible for the nonzero elements to form a cyclic group, and in fact this is the case for all finite fields.
It is very easy to come up with rings that aren't cyclic in either manner. Take, for example, $F\times F$ where F is the field of two elements. Its additive group is the Klein 4 group, and multiplivatively everything satisfies $x^2=x$, so there can't be any multiplicative generator of the set of three nonzero elements.

Let $R$ be a finite integral domain, and $a$ be a nonzero element of $R$. Why is it true that there are distinct $m, n$ such that $a^m = a^n$ in $R$?

Reading the lines of thought you included was helpful: they seem to suggest you are overthinking this.
The main way (already pointed out) is to note that if there weren't  powers of $a$ that coincided, the set of powers would be infinite, since there are as many powers as elements in $\Bbb N$.
To reframe it without a proof by contradiction, you could also use a Pigeon hole Principle argument. If the size of the ring is n, look at the first n+1 powers of a. You have to sort n+1 powers  into the n slots, and so one slot must contain at least two elements.
