Showing an infinite number of irreducible polynomials in $\Bbb Z_2 [x]$ exist Show that there are infinitely many irreducible polynomials in $\Bbb Z_2[x]$.

I have a strategy about how to do this but I need some help finding a contradiction. I want to assume there are finitely many irreducible polynomials. $p_1,...p_n$. Then I want to consider the irreducible factors of $(p_1p_2...p_n)+1$. I don't really know where to go from here. Any help would be appreciated.
$x+n$ is irreducible for all $n\in \mathbb Z$. They are also different irreducibles.

If question is about finite field with two element: each of root of your polynomial will be in some extension of $\mathbb F_2$. Think bigger extensions.
EDIT 2: Thanks everyone for downvoting a legitimate answer and an honest attempt to help. Great community here.

You can use Eisenstein's criterion to note that $x^n - 2$ is irreducible in $\mathbb{Z}_2[x]$ for every integer $n\ge 2$.

Also if you don't require your irreducible polynomials to be monic, then for any one irreducible polynomial $f$, $uf$ is also irreducible for any unit $u\in\mathbb{Z}_2[x]$. If you know that every element of $\mathbb{Z}_2$ can be expressed as a power series $\sum_{k\ge 0}a_k 2^k$ where $a_k\in\{0,1\}$, then by Hensel's lemma every power series with $a_0 = 1$ corresponds to a unit in $\mathbb{Z}_2$. Thus $\mathbb{Z}_2$ has infinitely many units, so the family $\{uf : u\in\mathbb{Z}_2^\times\}$ is already an infinite family of irreducible polynomials (though not monic).

EDIT: If by $\mathbb{Z}_2$ you mean the field $\mathbb{F}_2$ with two elements, then your strategy is probably easiest. If $p_1,\ldots,p_n$ denote all the primes of $\mathbb{F}_2[x]$, then none of the $p_i$'s can divide $(p_1p_2\cdots p_n)+1$, since if say $p_1$ divides $(p_1p_2\cdots p_n)+1$, then $(p_1p_2\cdots p_n)+1 = p_1q$ for some $q\in\mathbb{F}_2[x]$, so
$$1 = p_1q - (p_1p_2\cdots p_n)$$
so
$$1 = p_1(q - (p_2p_3\cdots p_n))$$
which is to say that $p_1$ is a unit (ie, $p_1$ is invertible), which of course cannot be the case if $p_1$ is prime/irreducible (these are the same in $\mathbb{F}_2[x]$ because $\mathbb{F}_2[x]$ is a UFD)