I'm new to the finite field theory, however after doing some trivial search on primitive polynomials, it seems that the polynomials of the form $$x^{2\cdot3^n}+x^{3^n}+1 \pmod 2$$ are irreducible.
Is it possible to prove it? I think it isn't possible to apply the method of checking all of the generators ($\phi(2^{2\cdot 3^n}-1)$ ) to find the order of the zeros (see this question). What should I do?
It is easy to show that this polynomial is not primitive.
I checked the claim for $n=\{1,2,3,4,5,6\}$ using WolframAlpha and it is true.
Notice that $(x^{2\cdot 3^n}+x^{3^n}+1)(x^{3^n}-1)=x^{3^{n+1}}-1$, so your polynomial is the $3^{n+1}$st cyclotomic polynomial.
The $n$th cyclotomic polynomial is irreducible over $\mathbb{F}_p$ exactly when $p$ is a primitive root mod $n$.
Since 2 is always a primitive root mod $3^n$, your polynomial is irreducible.