$X^n-Y^m$ is irreducible in $\Bbb{C}[X,Y]$ iff $\gcd(n,m)=1$

Geometric solution.

Consider the morphism of affine varieties $\mathbb{A}^1 \to V(X^n-Y^m), t \mapsto (t^m,t^n)$. It is surjective: If $x^n=y^m$ in $\mathbb{C}$, wlog $x,y \in \mathbb{C}^*$, choose some $t \in \mathbb{C}^*$ such that $x=t^m$. Then $y^m=x^n=t^{mn}$, hence $y=t^n \cdot \zeta$ for some $\zeta \in U(m)$. Since $m,n$ are coprime, $\zeta=\eta^n$ for some $\eta \in U(m)$. Then $t \eta$ is a preimage of $(x,y)$.

It follows that $V(X^n-Y^m)$ is irreducible, i.e. $\sqrt{(X^n-Y^m)}$ is a prime ideal. But the ideal $(X^n-Y^m)$ is radical, because $X^n-Y^m$ is square-free, since $\partial_X (X^n-Y^m)=n X^{n-1}$ and $X^n-Y^m$ are coprime in $\mathbb{C}(Y)[X]$.

Assume $f(X,Y) =X^n-Y^m=g(X,Y)h(X,Y)$. Then $f(Z^m,Z^n)=0$ implies that one of $g(Z^m,Z^n)$ or $h(Z^m,Z^n)$ is the zero polynomial. Suppose that $g(Z^m,Z^n)=0$. That means that for all $k$, the monomials cancel, i.e. if $$g(X,Y)=\sum a_{i,j}X^iY^j $$ then $$\sum_{mi+nj=k}a_{i,j}=0.$$ Can we ever have $mi+nj=mi'+nj'$? That would mean $m(i-i')=n(j-j')$, hence $n\mid i-i'$ (because none of $n$'s prime factors are in $m$) and likewise $m\mid j-j'$. So if $i>i'$ this implies $j> j'$.

I think Hagen von Eitzen's proof is good but not as explicite at the end.

I'll try to clarify:

"Assume $f(X,Y) =X^n-Y^m=g(X,Y)h(X,Y)$. Then $f(Z^m,Z^n)=0$ implies that one of $g(Z^m,Z^n)$ or $h(Z^m,Z^n)$, wlog $g(Z^m,Z^n)$, is the zero polynomial. That means that for all $k$, the monomials cancel, i.e. if $$g(X,Y)=\sum a_{i,j}X^iY^j $$ then $$ \sum_{mi+nj=k}a_{i,j}=0 $$

Can we ever have $mi+nj=mi'+nj'$? That would mean $m(i-i')=n(j'-j)$. It follows that $i=i'(\bmod n)$ and $j=j'(\bmod m)$. But $i,i'\in {0,1,...,n-1}$ and $j,j'\in {0,1,...,m-1}$. This implies $i=i'$ and $j=j'$."

Overall, $mi+nj=mi'+nj'$ is echivalent with $i=i'$ and $j=j'$. So distinct terms from $g$ and $h$ corespond to distinct terms in $g(Z^m,Z^n)$ and $h(Z^m,Z^n)$. So when we substitute $X=Z^m$ and $Y=Z^n$ none of the terms $g(Z^m,Z^n)$ and $h(Z^m,Z^n)$ will cancel out so that $g(Z^m,Z^n)=0$ and $h(Z^m,Z^n)=0$. That means $g(Z^m,Z^n)\ne0\ne h(Z^m,Z^n)$, but $0=f(Z^m,Z^n)=g(Z^m,Z^n)h(Z^m,Z^n)\ne0$, contradiction.