$X^n-Y^m$ is irreducible in $\Bbb{C}[X,Y]$ iff $\gcd(n,m)=1$ 
I am trying to show that $X^n-Y^m$ is irreducible in $\Bbb{C}[X,Y]$ iff $\gcd(n,m)=1$ where $n,m$ are positive integers. 

I showed that if $\gcd(n,m)$ is not $1$, then $X^n-Y^m$ is reducible. How to show the other direction. Please help.

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 $\mathbb{C}(Y)[X]$.
