I know that $ax=1$ has a solution in $F$ so that every element must be a unit but then I'm not sure how to proceed.
Hint $\rm\rm\,\ x \; f(x) = 1 \,$ in $\rm\ F[x]\, \Rightarrow\, 0 = 1 \, $ in $\rm F \, $ by evaluating at $\rm\ x = 0.\ $
If you know the universal (mapping) property of the polynomial ring then you may find it instructive to interpret the above from that viewpoint (see here).
It's obvious that $F[x]/(x) $ is isomorphic to $F$, and hence $(x)$ is a non trivial proper ideal of $F[x]$, and hence $F[x]$ can't be a field.
(Note that there are other trivial ways of doing this problem, as mentioned in the comments, and an answer above, but I thought of doing this problem in a bit different way, just for fun.)