Below are two noteworthy variations on Euclid's classic proof that there are infinitely many primes. The first is a simplification and the second is a generalization to rings with few units (= invertibles).
Theorem $\rm\, \ N (N+1) \,$ has a larger set of prime factors than does $\,\rm N > 0$.
Proof $\ $ $\rm N+1 > 1\,$ so it has a prime factor $\rm P$ (e.g. its least factor $> 1)$. $\,\rm N$ is coprime to $\rm N+1\,$ so $\rm P$ can't divide $\rm N$ (else $\rm\, P$ divides $\rm N+1\,$ and $\rm N$ so also their difference $\rm N+1 - N = 1).\,$ So the prime factors of $\rm\, N(N+1)$ include all those of $\rm N$ and at least one prime $\rm P$ not dividing $\rm N$.
Corollary $\ $ There are infinitely many primes.
Proof $\, $ Iterating $\rm\, N\to N (N+1)\, $ yields integers with an unbounded number of prime factors.
Below, generalizing Euclid's classic argument, is a simple proof that an infinite ring
has infinitely many maximal (so prime) ideals if it has fewer units than elements
(i.e. smaller cardinality). The key idea is that Euclid's construction of a new prime
generalizes from elements to ideals, i.e. given some maximal ideals $\rm P_1,\ldots,P_k$
then a simple pigeonhole argument employing $\rm CRT$ implies that $\rm 1 + P_1\cdots P_k$
contains a nonunit, which lies in some maximal ideal $\rm P$ which, by construction,
is comaximal (so distinct) from the prior max ideals $\rm P_i.\,$ Below is the full proof, excerpted from from some of my old sci.math/AAA/AoPS posts.
Theorem $\ $ An infinite ring $\rm R$ has infinitely many maximal ideals
if it has fewer units $\rm U = U(R)$ than it has elements, i.e. $\rm\:|U| < |R|$.
Proof $\rm\ \ R$ has a max ideal $\rm P_1,\:$ since the nonunit $\rm\: 0\:$ lies in some max ideal.
Inductively, suppose $\rm P_1,\ldots,P_k$ are maximal ideals in $\rm R$, with product $\,\rm J.$
$\rm Case\ 1\!: \; 1 + J \not\subset U.\:$ Thus $\rm 1 + J$ contains a nonunit $\rm p,\,$ lying in a max
ideal $\rm P.$
It's new: $\rm\: P \neq P_i\:$ since $\rm\: P + P_i = 1\:$ via $\rm\: p \in P,\ 1 - p \in J \subset P_i$
$\rm Case\ 2\!: \; 1 + J \subset U$ is impossible by the following pigeonhole argument.
$\rm R/J = R_1 \times \cdots \times R_k,\ R_i = R/P_i\:$ by the Chinese Remainder Theorem.
We deduce that $\rm\ |U(R/J)| \leq |U|\ $ because $\rm\ uv \in 1 + J \subset U \Rightarrow u \in U.$
Thus $\rm|U(R_i)| \leq |U(R/J)| \leq |U|\:$ via the injection $\rm u \mapsto (1,1,\ldots,u,\ldots,1,1).$
$\rm R_i$ field $\rm\: \Rightarrow\ |R| > 1 + |U| \geq |R_i|,\,$ and $\,\rm|J| \leq |U| < |R|\,$ via $\,\rm 1 + J \subset U.$
Therefore $\rm|R| = |R/J|\ |J| = |R_1|\ \cdots |R_k|\ |J|\:$ yields the contradiction that
the infinite $\rm|R|$ is a finite product of smaller cardinals.
I recall the pleasure of discovering this "fewunit" generalization of Euclid's proof and other related theorems while reading Kaplansky's classic textbook Commutative Rings as an MIT undergrad. There Kaplansky presents a simpler integral domain
version as exercise $8$ in Section $1$-$1,\:$ namely
(This exercise is offered as a modernization of Euclid's theorem on
the infinitude of primes.) Prove that an infinite integral domain with
with a finite number of units has an infinite number of maximal ideals.
I highly recommend Kap's classic textbook to anyone seeking
to master commutative ring theory. In fact I highly recommend
everything by Kaplansky - it is almost always very insightful and
elegant. Learn from the masters! For more about Kaplansky see
this interesting NAMS paper which includes quotes from many eminent
mathematicians (Bass, Eisenbud, Kadison, Lam, Rotman, Swan, etc).
I liked the algebraic way of looking at things.
I'm additionally fascinated when the algebraic
method is applied to infinite objects.
$\ $--Irving Kaplansky
Note $ $ Readers familiar with the Jacobson radical may note that it may be employed to describe the relationship between the units in $\rm R$ and $\rm R/J\:$ used in the above proof. Namely
Theorem $\ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R).$
$\rm(1)\quad J \subseteq Jac(R),\quad $ i.e. $\rm\:J\:$ lies in every max ideal $\rm\:M\:$ of $\rm\:R$
$\rm(2)\quad 1+J \subseteq U,\quad\ \ $ i.e. $\rm\: 1 + j\:$ is a unit for every $\rm\: j \in J$
$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad\ $ i.e. proper ideals survive in $\rm\:R/J$
$\rm(4)\quad M\:$ max $\rm\:\Rightarrow M+J \ne 1,\quad $ i.e. max ideals survive in $\rm\:R/J$
Proof $\: $ (sketch) $\ $ With $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$
$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.
$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow I = 1$
$\rm(3\Rightarrow 4)\ \ \ $ Let $\rm\:I = M\:$ max.
$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\:$ by $\rm\:M\:$ max.