I would like to show that the polynomial $y-x^3$ is irreducible in $F[x,y]$, where $F$ is an arbitrary field. Here is my attempt.
Since $y-x^3$ is a linear polynomial in $y$ with coefficients in $F[x]$, then if it were reducible, we would be able to express it as a product $y-x^3 = c f(y)$, where $c$ is a constant non-unit coefficient in $F[x]$ and $f$ a linear polynomial.
But then $c$ must also divide $x^3$, so that $c$ is either some power (up to degree 3) of $x$ or otherwise a unit in $F[x]$. It cannot be the former, as $c$ must also divide $y$ and both $x, y$ are irreducible in $F[x,y]$. It also cannot be the latter, as by choice $c$ must be a non-unit. Thus $y-x^3$ is irreducible.
Assuming this is okay, I am also concerned about if I have correctly not used any underlying assumptions about the field $F$. I am only really familiar with fields like $\mathbb{C}$, $\mathbb{Q}$, and so on, and I am not sure if somewhere implicitly I have required the characteristic to be zero, or to be algebraically closed, or anything else. Some confirmation here would be ideal too.
Here's another approach, which behind the scenes works a lot like your work, but by using irrationals can do much more "at once":
Pick a positive irrational number $\alpha$ and for $f(x,y)=\sum_{n,m}a_{n,m}x^ny^m$, define $$\deg_\alpha f=\max\{\,n+\alpha m\mid a_{n,m}\ne0\,\}. $$ Then, just as with "usual" degrees, we have $\deg_\alpha(fg)=\deg_\alpha f+\deg_\alpha g$. The key point is but that for irrational $\alpha$, the map $(n,m)\mapsto n+\alpha m$ is injective.
Now assume $$y-x^3=f(x,y)g(x,y).$$ So if $\deg_\alpha f=n+m\alpha$ and $\deg_\alpha g=r+s\alpha$ with $n,m,r,s\in\Bbb N_0$, then $$ (n+r)+(m+s)\alpha=\deg_\alpha(y-x^3)=\max\{\alpha,3\}.$$ If we take $\alpha=\pi$ or $\alpha=\sqrt{10}$, say, wo obtain $\alpha$ on the right and conclude that $n+r=0$ and $m+s=1$, i.e., $n=r=0$ and one of $m,s$ is also $=0$, i.e., one of $f,g$ is constant.