Proving that $\langle x, y \rangle$ is a not principal ideal of $\mathbb{Q}[x, y]$ My idea was to prove that $x, y$ aren't elements of this ideal, the way I thought about it is that being part of this ideal forces $x$ to be written in the form of a product between $2$ polynomials in $\mathbb{Q}[x, y]$, but $1$ of those polynomials will have to be $x$ and the other will have to be $1$, the unit, which isn't possible... I don't know how to formulate this correctly.
Could someone please help me with this question?
Your idea is good. In fact, assuming the ideal is generated by some polynomial $P$, we deduce the following :

The degree in $y$ of $P$, when $P$ is seen as an element of $\mathbb{Q}[x][y]$, has to be $0$, because $x$ is in the ideal generated by $P$. Can you explain why ?

Note that $A:=\mathbb{Q}[x]$ is a domain, so we can work with polynomials with coefficients in $A$ without much trouble.

Similarly, the degree in $x$ of $P$ has to be $0$.

So $P$ is constant.
So, the ideal generated by $P$ is either the whole ring, or the $0$ ideal. In both cases, it is not the ideal $(x,y)$. By contradiction, the result is proved.
Suppose $x=f(x,y)$ and $y=f(x,y)$ for some polynomial $f(x,y)$.
In view of $x=f(x,y)$, write $f(x,y)$ as a polynomial in $x$ with coefficients given by polynomials in $y$ such as in $x^2(y+1) + x(y-1) + 1$.
Then $x= x^n f_n(y) +\ldots +xf_1(y) + f_0$. By comparing degrees, $n=1$ and then $f_1(y)=1$ by the shorting rule, i.e., $f(x,y)=x$. Thus there is no other representation of the polynomial $x$.
Similar for $y=f(x,y)$.