Is the number of irreducibles in any number field infinite?

Are there infinitely many irreducibles in the ring of integers of any algebraic number field ?

I tried to follow the same argument as we usually do for integers. Suppose there are finitely many irreducibles, say $p_1,\ldots ,p_n$ and let $\alpha :=1+ p_1\cdots p_n$. Now if $\alpha $ is not a unit then it must have an irreducible $p$ such that $p|\alpha$ but then $p$ can not be any of the $p_i$'s and we have a contradiction. $\textit{What if $\alpha$ is an unit ? Is it possible for $\alpha$ to be an unit ?}$

Of course, one can replace $p_1\cdots p_n$ by $p_1^{k_1}\cdots p_n^{k_n}$ for any $k_1,\ldots ,k_n\in\mathbb{N}$ and the same argument would go through.

$\alpha$ can be a unit as the element $1+\sqrt{2}$ in the ring of integers $\mathbb{Z}[\sqrt{2}]$ shows. Here's a slight variation of your argument: Let $p_1,\ldots,p_n$ be all the irreducibles. Since $x=p_1\cdot\ldots\cdot p_n$ is integral over $\mathbb{Z}$ it satisfies: $$ 0\neq-a_0=x^n+\ldots+a_1x=x(x^{n-1}+a_{n-1}x^{n-2}+\ldots+a_1)=x\cdot y\in\mathbb{Z}. $$ Taking the negative of $y$ if necessary wlog we assume $xy>0$. Then $1+xy$ is not a unit of $\mathbb{Z}$ (and so also not a unit of the ring of integers $R$). Now you can conclude with your argument.

Consider the mapping

$$\iota \colon \mathbb{P} \to \mathfrak{P}(R);\quad \iota(p) = \lbrace a \in R \colon a \mid p \land a \text{ is irreducible}\rbrace,$$

where $R$ is the ring under consideration, and $\mathbb{P}$ is the set of (positive) rational primes.

Since the norm of the field (w.r.t. $\mathbb{Q}$) is integer-valued on $R$, and multiplicative, each division chain from a nonzero element must stop (at an irreducible element), and hence $\iota(p) \neq \varnothing$ for all $p$.

Let $z \in \iota(p) \cap \iota(q)$. Then $z \mid \gcd_{\mathbb{Z}}(p,\,q)$. If $p \neq q$, then $\gcd_{\mathbb{Z}}(p,\,q) = 1$, hence $\iota(p) \cap \iota(q)$ contains only units - but $\iota(p)$ contains no units by definition. Thus $p\neq q \Rightarrow \iota(p) \cap \iota(q) = \varnothing$. $\iota(p)$ is closed under $x \sim y \iff \bigl(\exists \varepsilon \in R^\ast\bigr)(y = \varepsilon\cdot x)$.

Hence for any choice function $c$,

$$c \circ \iota \colon \mathbb{P} \to R$$

is injective, and $\lbrace c(\iota(p)) \colon p \in \mathbb{P} \rbrace$ is an infinite family of pairwise non-associated irreducible elements of $R$.