Proving ring $R$ with unity is commutative if $(xy)^2 = x^2y^2$ 
Let $R$ be a ring with unity. Assume that $\left(xy\right)^2 = x^2 y^2$ for all $x, y \in R$. Prove that $R$ is commutative.

I tried several methods to solve this but couldn't get through. Now the solution in almost all the textbooks goes like this.
First take $x$ and $y+1$ so that $ (x(y+1))^2 = x^2(y+1)^2 => xyx = x^2y \\$ 
Then substitute $x+1$ in place of $x$ to get $yx = xy$
I have always believed that solutions to questions can't be abrupt, there has to be a logic behind approaching in a particular way. Therefore, I couldn't grasp the intuition behind this method.
Can somebody please suggest some other intuitive way or explain the reasoning behind above solution?
 A: The idea is the following: It suffices to consider the subring generated by two arbitrary elements $x,y$ (since if this is commutative, then in particular $xy=yx$ and we are done). Now, it is not true that $(xy)^2=x^2 y^2$ implies $xy=yx$. But actually we have a stronger condition, namely that $(pq)^2=p^2 q^2$ holds for all (non-commutative) polynomials in $x$,$y$. The simplest non-constant polynomials are linear factors $x+ z$ and $y + z$ (with $z \in \mathbb{Z}$). That's why one chooses them first and tries to combine the calculations for various $z$.
PS: It is really unfortunate that many people (are forced to) write down proofs without mentioning any ideas of the proof. From the ideas one can really benefit. The proof alone is just a bunch of sentences even a computer can check for validity.
A: By Dan Shved's comments, you can see why switching $\Box$ by $\Box +1$ is a commen way in commutativity conditions. The commutator $[x,y]$ plays an vital role in commutativity conditions. There is immense number of papers which are about this technique. For example:
1) I. N. HERSTEIN, Two remarks on the commutativity of rings, Canad,J. Math.,7 (1955),411~412.
2)E. C. JOHNSEN, D. C. OUTCALT and A. YAQUB, An elementary commutativity therem for rings,
Amer. Math. Monthly, 75 (1968), 288--289.
3)Khan, On Commutativity Theorems for Rings, Southeast Asian Bulletin of Mathematics (2002) 25: 623–626.
Also, using nilpotent elements is the another technique for commutativity conditions.
A: This is probably not the best possible answer, but still, I think it can make things a bit clearer.
Let's begin by taking the given identity and subtracting parts from each other. We get this: $$x[x, y]y = 0,$$
where $[x, y] = xy - yx$. This is not what we want: we want $[x, y] = 0$. If $R$ were a division ring, we would already be done (just divide by $x$ on the left and $y$ on the right). If $R$ didn't have zero divisors, we would also be done (it's allowed to remove non-zero factors from equations in such rings). But we have an arbitrary ring, so some trickery is required to get rid of the $x$ and $y$ around $[x, y]$.
Adding $1$ to one of the variables seems to move things forward: indeed, it is very easy to check that $[x+y, z] = [x, z] + [y, z]$ for all $x, y, z \in R$, so $[x+1, y] = [x, y] + [1, y] = [x, y]$. Thus, substituting $x+1$ for $x$ gives us
$$
(x+1)[x, y]y = 0.
$$
Subtracting this from the original equation, we get $[x, y]y = 0$. We managed to kill $x$ on the left, even though there is no division in our ring! Now we do the same trick with $y$, and we have proved the desired $[x, y]=0$.
This doesn't explain where the idea of "adding $1$" comes from, but hopefully it makes it a bit less mysterious.
