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: 
Consider both sides of the equations as polynomials in $x$, and use the Fundamental Theorem of Algebra to factorise each side into linear factors of the form $(x-f\_k(y))$.

Note then that the multiplicity of the factor $(x+iy)$ is odd on the left-hand side (one factor from $(x^2+y^2)$, plus double its multiplicity in $p$), but odd on the right-hand side (double its multiplicity in $q$). So we have a contradiction, and the two expressions cannot be equal as polynomials.
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.