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.
Hint: Take the square root of both sides. There are some signs you need to think about, but basically can $\sqrt{x^2+y^2}p(x,y)$ ever be a polynomial for $p\ne 0\,?$