Explicit large finite fields in characteristic $2$

The following paper provides an explicit computational approach to your problem:

J. D. Swift: Construction of Galois fields of characteristic two and irreducible polynomials, Math. Comput. 14, 99-103 (1960). ZBL0105.01202.

Nice question!

The case of self-reciprocal irreducible trinomials and pentanomials is of interest in cryptography.

The linked paper which I was not aware of before your question has the following result, see here

$$x^{3^s10}+x^{3^s9} +x^{3^s5}+x^{3^s}+1$$is a self-reciprocal irreducible pentanomial of degree $3^s10≡ 6 \pmod{12}$ for every positive integer $s$.

For $n=3^k$, the polynomial $p=x^{2n}+x^n+1$ is irreducible over $\mathbb{F}_2$.

Proof: By Rabin's irreducibilty test, it suffices to check that $p|x^{2^{2n}}-x$ and $\gcd(p,x^{2^{2n/3}}-x)=1$.

Note that the order of $x$ mod $p$ is $3n=3^{k+1}$. Hence, since $3^{k+1}|4^{n}-1$ by lifting-the-exponent lemma, we have $p|x^{2^{2n}-1}-1$.

Again by lifting-the-exponent lemma, we have $4^{3^{k-1}}-1=3^km$ for $m$ not divisible by $3$. Hence $x^{4^{3^{k-1}}-1}=x^{nm}\equiv x^n\ \text{or}\ x^{2n} \pmod{p}$ since $x^n$ has order $3$. As $(x^n-1)x^n=(x^{2n}-1)x^{2n}=1$, this mean that $x^{2^{2n/3}-1}-1$ is invertible in $\mathbb{F}_2[x]/(p)$. As $x$ is also invertible, we have shown that $\gcd(p,x^{2^{2n/3}}-x)=1$, as wanted.