= Solution
A monic <quadratic polynomial> over a <field> is irreducible exactly when it has no root in that <field>. Testing the nine choices of its two coefficients in $\mathbb F_3$ gives
$$
\boxed{x^2+1,\qquad x^2+x+2,\qquad x^2+2x+2.}
$$
For clarity, a zero constant term gives a root at zero. Among constant term one, only the zero linear coefficient avoids roots at $1,2$. Among constant term two, precisely the linear coefficients one and two avoid roots. This exhausts all choices.
Put $f=x^4+x+2$. Its values at $0,1,2$ are respectively $2,1,2$, so it has no linear factor. If this quartic were reducible, absence of a linear factor would force a factorization into two irreducible quadratics. Division by the three candidates above gives remainders respectively $x,x+1,x+1$, all nonzero. Thus \b[$f$ is an <irreducible polynomial>].
The <quotient by an irreducible polynomial is a field> consequently gives a <field> of $3^4=81$ elements. Its nonzero elements form a <multiplicative group of a finite field> of order $80$. In this <field>, $a^4=1+2a$. Successive squaring gives
$$
a^8=1+a+a^2,\qquad a^{16}=2+a+2a^3.
$$
The latter is not one: otherwise the nonzero <polynomial> $1+x+2x^3$ of degree below four would vanish at $a$, contradicting its <minimal polynomial>. By <Lagrange's theorem>, $(a^{16})^5=a^{80}=1$. Since five is prime, its order is exactly five. Hence one answer is
$$
\boxed{\zeta=2+a+2a^3.}
$$
Its other nonidentity powers are the remaining primitive fifth roots. It is unnecessary to prove that $a$ itself generates the entire multiplicative <group>.
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