First, let me explicitly assume $R$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$.
Proposition. The Krull dimension of $R$ is at most $1$.
Proof. Let $\mathfrak{m} = (t)$, and let $\mathfrak{p}$ be prime. We know $\mathfrak{p} \subseteq \mathfrak{m}$, so it is enough to show that either $\mathfrak{p} = \mathfrak{m}$ or $\mathfrak{p} = (0)$. Suppose $\mathfrak{p} \ne \mathfrak{m}$: then $t \notin \mathfrak{p}$. Let $a_0 \in \mathfrak{p}$. For each $a_n$, because $\mathfrak{p}$ is prime, there exists $a_{n+1}$ in $\mathfrak{p}$ such that $a_n = a_{n+1} t$. By the axiom of dependent choice, this yields an infinite ascending sequence of principal ideals:
$$(a_0) \subseteq (a_1) \subseteq (a_2) \subseteq \cdots$$
Since $R$ is noetherian, for $n \gg 0$, we must have $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$. Suppose, for a contradiction, that $a_0 \ne 0$. Then, $a_n \ne 0$ and $a_{n+1} \ne 0$, and there is $u \ne 0$ such that $a_{n+1} = a_n u$. But then $a_n = a_{n+1} t = a_n u t$, so cancelling $a_n$ (which we can do because $R$ is an integral domain), we get $1 = u t$, i.e. $t$ is a unit. But then $\mathfrak{m} = R$ – a contradiction. So $a_n = 0$. $\qquad \blacksquare$
Here's an elementary proof which shows why we can reduce to the case where $R$ is an integral domain.
Proposition. Any non-trivial ring $A$ has a minimal prime.
Proof. By Krull's theorem, $A$ has a maximal ideal, which is prime. Let $\Sigma$ be the set of all prime ideals of $A$, partially ordered by inclusion. The intersection of a decereasing chain of prime ideals is a prime ideal, so by Zorn's lemma, $\Sigma$ has a minimal element. $\qquad \blacksquare$
Thus, we can always assume that a maximal chain of prime ideals starts at a minimal prime and ends at a maximal ideal. But if $R$ is a noetherian local ring with principal maximal ideal $\mathfrak{m}$ and $\mathfrak{p}$ is a minimal prime of $R$, then $R / \mathfrak{p}$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$, and $\dim R = \sup_\mathfrak{p} \dim R / \mathfrak{p}$, as $\mathfrak{p}$ varies over the minimal primes.
Update. Georges Elencwajg pointed out in a comment that the first proof actually works without the assumption that $R$ is a domain, because $(1 - u t)$ is always invertible.
Exercise: a local von Neumann regular ring with identity is a division ring.
Proof: If $a$ is nonzero, $axa=a$ for some element $x\in R$, and $ax$ and $xa$ are nonzero idempotents. Since local rings only have trivial idemoptents, $ax=1=xa$. Thus every nonzero element is invertible.
For the purposes of the question, I suppose that "local" for a ring without identity means "has one proper ideal containing all proper ideals." Along with von Neumann regularity, this is enough to prove $R$ has an identity.
Suppose $R$ is VNR and $M$ is the unique maximal ideal. Select $a\in R\setminus M$. Then $a=axa$ and $ax=e$ is an idempotent. Clearly $ax\notin M$, for if it were, $axa=a\in M$.
Then $(e)$ is some ideal of $R$. Suppose $(e)\neq R$: then by our supposition of what "local" means above, $(e)\subseteq M$, contradicting $e\notin M$.
Therefore the only possibility is that $(e)=eR=R$, but then it is easy to see that $e$ is the multiplicative identity of $R$. At that point, the exercise above shows $R$ is a field.
It works with minor modifications in the noncommutative case. Now we suppose there is a proper right ideal containing all proper right ideals, and a proper left ideal containing all proper left ideals.
Use $e=ax$ and $e'=xa$ to argue the same way, and you wind up with $eR=R=Re'$ to get that $e$ is a left identity and $e'$ is a right identity, therefore $e=e'$ is the identity for the ring. The exercise above indicates $R$ is a division ring.
Best Answer
The characteristic of a local ring is a power of a prime or $0$, and any of these happens in some local rings.
That they all happen is easy : you may look at fields for characteristic $0$, and $\mathbb{Z}/p^n\mathbb{Z}$ for powers of primes.
Now let $(R,m)$ be a local ring, and $n$ its characteristic, which we assume to be $>0$. Suppose $n=ab, a\land b = 1$. Then the ideals $I=\{x\in R, ax = 0\}$ and $J=\{x\in R, bx=0\}$ are comaximal : indeed $a\in J, b\in I$ and there are $u,v$ with $au+bv=1$ so $1\in I+J$.
Therefore by locality, one of them is $R$ (otherwise they would both be $\subset m$). If it is $I$, then $a = 0$ in $R$ and so $R$ has characteristic $\mid a$ so $b=1$. If it's $J$, then $a=1$. In any case, $a=1 \lor b=1$, so that $n$ is a power of a prime.