Is there a way to find one irreducible polynomial of degree n in the field Z2

If $p(x)$ is is a prime polynomial, then it is an irreducible polynomial. If $p(x)$ is is a prime polynomial of degree $n$, then $\mathbb{Z}[x]/p(x)$ is a field of order $2^n$, and so $p(z)|z^{2^n}-z$. This will greatly narrow down the searching.

Most techniques rely on things specific to the exact degree unfortunately, and there is no useful general technique AFAIK.

In general it is difficult, but since testing irreducibility is reasonably fast (for a suitable definition of reasonably fast) finding one by random poking won't take too long because a random degree $n$ polynomial is irreducible with probability approximately $1/n$.

As Stella Biderman said, most methods are specific to the degree, and consequently ad hoc. Degree 20 you said? That is easy, because

• $20=\phi(25)$, and
• $2$ generates the group $\Bbb{Z}_{25}^*$.

These two facts together imply that the characteristic zero cyclotomic polynomial $$ \Phi_{25}(x)=\frac{x^{25}-1}{x^5-1}=x^{20}+x^{15}+x^{10}+x^5+1 $$ remains irreducible over $\Bbb{Z}_2$.

This is seen as follows. Let $\alpha$ be a primitive root of unity of order $25$ (in some extension field of $\Bbb{Z}_2$). Let $m(x)$ be the minimal polynomial of $\alpha$. By Galois theory we get all the zeros of $m(x)$ by repeatedly applying the Frobenius automorphism, $F:x\mapsto x^2$, to $\alpha$, so the zeros of $m(x)$ include $$ \alpha,\alpha^2,\alpha^4,\alpha^8,\alpha^{16},\alpha^{32}=\alpha^7, \alpha^{14},\ldots $$ By the second bullet above the list contains exactly the powers $\alpha^k, 1\le k<25, \gcd(k,25)=1$. In other words all the primitive roots of order $25$. Therefore $$ m(x)=\Phi_{25}(x) $$ is irreducible modulo two.