Suppose $R$ is a finite ring (commutative ring with $1$) of characteristic $3$ and suppose that for every unit $u \in R\:,\ 1+u\ $ is also a unit or $0$. We need to show that $R$ is a field. Is this true if ${\rm char}(R) > 3$?

Here is what I attempted to do. $\:$ First of all, $\:$ I noticed that the statement is not true if $\ R\ $ is infinite ($ \mathbb F\_3[x]$ is an example of an infinite ring which is not a field but it satisfies all the required properties). Now, in a finite ring, a non-zero element is either a unit or a $\:0\:$ divisor, so I tried to show that $R$ has no $\:0\:$ divisors. Clearly, $R$ has no nonzero nilpotent elements (if $x$ is nilpotent, then $1+x$ is a unit, but then $1+(1+x)$ and $1+(2+x)$ is either a unit or $\:0\:.\:$ Hence $x$ is either a unit or $\:0\:,\:$ and since $x$ is nilpotent, it can't be a unit, so we must have $x = 0$). But this does not solve the problem, since $R$ could have elements that are $\:0\:$ divisors but not nilpotent (for example, $\ (1,0)\ $ is a $\:0\:$ divisor in $\ \mathbb Z/3\:\mathbb Z \times \mathbb Z/3\:\mathbb Z\ $ but it is not nilpotent).

Another observation I made is that the set of units, together with $\:0\:$ forms a group under addition, so that $J = R^{\*}$, together with $\:0\:$ is a subring of $R$. hence we may view $R$ as a $J$-module (and since $J$ is clearly a field, $R$ is a $J$-vector space). 

Another thing I tried is to show that $R$ has no proper nontrivial ideals. Viewing $R$ and $J$ as abelian groups, I noticed that a non-trivial ideal of $R$ can contain at most one element from each coset of $J$ in $R$, because if an ideal contains two distinct elements from the same coset of $J$ in $R$, this ideal would have to contain their difference, hence it'd have to contain a unit, hence it would not be a proper ideal. But again, I don't see how this observation leads to a solution.

As for the last part, I suspect that this statement will remain true if ${\rm char}(R) > 3$. Since $1$ is a unit, it follows that $1,2,3,\ldots$ are all either units or $0$, which can only happen if ${\rm char}(R) = p$, a prime number (and then I suspect that $R$ will have to be a finite field), but again I do not see how to prove (or disprove) this.

By the way, this is not a homework problem. I am studying algebra on my own, and after thinking about it for a few days and making the observations I listed above, I still don't see how to finish the proof. I would appreciate your suggestions. Thank you in advance.