Given a field $F$ and $f,g \in F$, I'm given the task of proving what it says in the title.
So far, I've only managed to notice that $\deg f>0$ and $\deg g>0$, since if at least one were a 'constant' then multiplying by the inverse of such element gives $1_F\in N$ and as such wouldn't be a proper ideal.
We can prove the converse. If $f$ and $g$ are irreducible with distinct degrees, then $F[x]=(f,g)=\{rf+sg\ :\ r,s\in F[x]\}$.
So suppose $f$ and $g$ are irreducible polynomials with distinct degrees. Then $f$ and $g$ are distinct so the only common factors that $f$ and $g$ share are the units in $F[x]$ ( so elements in $F$). Hence $\text{gcd}(f,g)=1$. Since $F$ is a field, there is a division algorithm in $F[x]$ so that there exists $r,s\in F[x]$ such that $rf+sg=1$. But this implies that $(f,g)=F[x]$.