For an integral domain $R$, the rings $R\times R $ and $R\times R\times R$ are not isomorphic 
For an integral domain $R$, the rings $R\times R $ and $R\times R\times R$ are not isomorphic

My attempt:
On contrary suppose that both are isomorphic then if G is prime ideal of one ring then its isomorphic copy must be prime ideal of other
we will construct projection map $p_1:R\times R\to R$
as $p_1(a,b)=a$
Now here kernel is $R$ 
then by prime ideal theorem $(R\times R )/R\cong R$ which is integral domain so R is prime ideal of R.
Now we will again construct projection map $p_2:R\times R\times R\to R\times R$
as $p_2(a,b,c)=(a,b)$
Now here kernel is $R$ Now $(R\times R \times R )/R\cong R\times R $ but it is easy to show that $R\times R $ is not integral domain.
So our assumption is wrong .
Hence both are not isomorphic.
Is my argument are valid?
Thanks in advanced.

Your argument is not valid: the quotients $(R\times R)/(0\times R)$ and $(R\times R\times R)/(0\times 0\times R)$ not being isomorphic does not rule out the existence of an isomorphism $\varphi\colon R\times R\to R\times R\times R$. It just rules out that such an isomorphism can satisfy $\varphi(0\times R)=0\times 0\times R$.

Let me clarify that for two isomorphic rings $R$ and $R'$ with ideals $I\subset R$ and $I'\subset R'$ that are also isomorphic as rings, you can not conclude that $R/I$ and $R'/I'$ are isomorphic rings:
Take any non-zero ring $S$ and let $R=\bigoplus_{i=1}^\infty S$ with component-wise addition and multiplication. Let $R'=R$, $I=R$ and
$$
I' = \left\{ \, (0,s_1,s_2,\dots)\in R \,\middle|\, s_i\in S\,\right\}.
$$
Note that the shift map
$$
(s_1,s_2,\dots) \mapsto (0,s_1,s_2,\dots)
$$
is a ring isomorphism from $I$ to $I'$.
However, $R/I=0$ while $R/I'\cong S$.
Also note that the shift map is no longer surjective when extended to a map $R\to R$
