Here is a quick and simple way to prove this, which does not depend on even/odd cases etc., or any of the (iteratively) linked citings, but is indeed quite concise and self contained:
As pointed out by our OP Ahmed in the text of the question itself, the "$\Rightarrow$" direction is virtually self-evident, that is, if $R$ has no non-zero nilpotents and $a^2 = 0$, then clearly $a = 0$ as well.
As for the "$\Leftarrow$" direction, we observe that
$a^2 = 0 \Longrightarrow a = 0 \tag 1$
implies, for any $x \in R$ and $m \ge 1$,
$x^{2^m} = 0 \Longrightarrow x^{2^{m - 1}} = 0, \tag 2$
since
$(x^{2^{m - 1}})^2 = x^{2^{m - 1}}x^{2^{m - 1}} = x^{2^{m - 1} + x^{2^m - 1}} = x^{2^1 2^{m - 1}} = x^{2^m} = 0. \tag 3$
In light of (2), we may evidently proceed downward, one decrement of $m$ at a time, until we eventually reach $m = 1$, viz:
$x^{2^m} = 0 \Longrightarrow x^{2^{m - 1}} = 0 \Longrightarrow x^{2^{m - 2}} = 0$
$\Longrightarrow \ldots \Longrightarrow x^2 = 0 \Longrightarrow x = 0. \tag 4$
Now if $b \in R$ is nilpotent,
$b^n = 0, \tag 5$
we simply choose $m$ sufficiently large that
$2^m \ge n; \tag 6$
then
$b^{2^m} = b^{2^m - n}b^n = 0; \tag 7$
at this point (2)-(4) take over and we conclude that
$b = 0; \tag 8$
$R$ has no non-zere nilpotent elements. $OE\Delta$.
You are way off in the weeds. This is especially the case because there is no justification for $c$ or $d$ to have an inverse.
The argument should simply be:
If $ab=0$ for nonzero $a$ and $b$, then by definition of primeness, $bRa\neq \{0\}$, so $bxa\neq 0$ for some $x$.
Then $bxa$ is a nonzero element whose square is zero.
Best Answer
Hint:
Start with, let $f \in R$ be nilpotent and use the fact "zero is the only nilpotent element in an integral domain" to conclude $f=0$