Question: Let $p(x, y)$ and $q(x, y)$ are real polynomials in $x, y\in \mathbb{R}$ and $p(x, y)\ne0$ and $q(x, y)\ne0$.

Why does $$(x^2+y^2)p(x, y)^2\ne q(x, y)^2$$ for all $x, y\in \mathbb{R}$

Answer: 
$\mathbb{R}[x,y]$ (the real polynomials) is what we call a *factorial ring* i.e. a ring where every $q\not=0$ can be decomposed uniquely as
$$
q=\lambda\prod\_{i\in I} p\_i^{\nu\_i(q)}
$$
where $\lambda$ is invertible (here is a non-zero real) and $p\_i$ is a complete family of irreducibles (see all standard texts). 
Now, $(x^2+y^2)$ is irreducible (one can choose it in the family) and 
$$
\nu\_{(x^2+y^2)}[(x^2+y^2)p(x,y)^2]=2n+1
$$  

whereas
$$
\nu\_{(x^2+y^2)}[q(x,y)^2]=2m
$$
with $n,m$ integers, a contradiction.
Assume equality holds, i.e.
$$(x^2 + y^2) p(x, y)^2 = q(x, y)^2$$
**Hint:** Can you make the LHS zero by choosing some $x$ and $y$? What would that mean for $q(x, y)$?

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EDIT: Assuming you meant $p \neq 0$ and $q \neq 0$ as polynomial, as JLA points out in the comments, the hint above is not correct. Here is a slight alternative of Duchamp Gérard H. E.'s answer:

$r(x, y) = x^2 + y^2$ is irreducible in $\mathbb{R}[x, y]$, so $q$ must have $r$ as a factor. Let $q = q\_1r$, then
$$p^2 = rq\_1^2.$$
Similar argument as before $p$ must have $r$ as a factor. Let $p = p\_1r$, and we get
$$rp\_1^2 = q\_1^2,$$
an equation similar to the original, but $\mathrm{degree}(p\_1) < \mathrm{degree}(p)$. This gives an [infinite descent](https://en.wikipedia.org/wiki/Proof_by_infinite_descent) on the degree, a contradiction.