Let $R$ be a ring without zero divisors then show that all nonzero elements of $R$ have same additive order 
Question Let $R$ be a ring without zero divisors then show that all nonzero elements of $R$ have same additive order.

(Though In the book the question is asked only for only for commutative ring but, I think we can prove the result for all rings! Am i correct?)
My attempt:
Let $x, y$ be any two nonzero elements of a ring $R$
Case(a): let $|x|$ (additive order) is infinite and suppose the (contrary) that, $|y|$ is finite say, $|y|=n$. Then,
$$\begin{align*}
0&= (ny)x\\ 
&=(y+\cdots+y\quad\text{(}n\text{ times)})x\\
&=yx+yx+\cdots+yx\quad(\text{(}n\text{ times)})\\
&=y(x+\cdots+x\quad(\text{(}n\text{ times)})\\
&=y(nx)
\end{align*}$$
Since $y≠0$ and $R$ has no zero divisors we must have, $nx=0$ which contradicts the fact that, $|x|$ is infinite.
Thus, our assumption must be wrong! and so we must have $|y|$ is infinite. Thus, we had proved that, if $R$ has atleast one nonzero element of infinite order then every nonzero element of $R$ has infinite order.
Case(b): let $|x|=n, |y|=m$ where $n<m$ then,
$$\begin{align*}
0&= (nx)y
\\ &=(x+\cdots+x\quad\text{(}n\text{ times)})y
\\&=xy+xy+\cdots+xy\quad(\text{(}n\text{ times)})
\\&=x(y+\cdots+y\quad(\text{(}n\text{ times)})
\\&=x(ny)
\end{align*}$$
As $x≠0$ and $R$ has no zero divisors we must have, $ny=0$ which contradicts the fact that, $|y|=m$ (since  $n<m$). Thus, our assumption that $|x|=n, |y|=m$ where $n<m$ is wrong!
So we must have $|x|=|y|$. Thus, if $R$ has atleast one nonzero element of finite additive order then all nonzero element have finite additive order and all of them have same order.
So by case(a) and case(b). We have, If $R$ is a ring without zero divisors then all nonzero elements of $R$ have same additive order.
Am I correct? Is there are any mistakes in proof?
Please help.
 A: Since there was some discussion about the noncommutative case in the comments, I thought I may as well include an alternative perspective on things which should work for noncommutative case as well.
The characteristic of a ring $R$ is defined to be $\text{char} R=\inf \{n\in \mathbb{Z}_{>0}:n\cdot 1_R=1_R\cdot n:=\underbrace{1_R+...1_R}_{n\text{ terms}}=0\}$
(It's not so important for us here but if you don't have zero divisors, the characteristic is a prime number or infinite).
Rather than comparing the additive order of arbitrary pairs of elements in $R$, we will show that the additive order of a nonzero element will be the same as the additive order of $1_R$ (i.e. the characteristic of the ring).
$\underbrace{r+...+r}_{n\text{ terms}}=r(\underbrace{1_R+...1_R}_{n\text{ terms}})=0$ when $n=\text{char} R$. So, the additive order of $r\in R$ is at most $\text{char}R$.
Conversely, if $n$ is the additive order of $r$, the same algebra together with the fact that there are no zero divisors in $R$ tell us the reverse inequality.
i.e. $\underbrace{r+...+r}_{n\text{ terms}}=0\implies r\cdot (\underbrace{1_R+...1_R}_{n\text{ terms}})=0 $ and since $r\neq 0$, $\underbrace{1_R+...1_R}_{n\text{ terms}}=0$ so $\text{char} R$ is at most the additive order of $r\in R$.
Actually since $1_R$ commutes with everything even in the noncommutative case, we only need that there are no left (or right) zero divisors in $R$!
