Zeroes of irreducible two variables polynomial which is reducible in algebraic closure 
Let $\mathbb{F}$ be a field of characteristic $0$, $\bar{\mathbb{F}}$ its algebraic closure, $p(x,y) \in \mathbb{F}[x,y]$ an irreducible polynomial which is reducible in $\bar{\mathbb{F}}[x,y]$. Show that $p(x,y)$ has only a finite number of zeroes in $\mathbb{F}^2$.

The approach suggested involves demonstrating something like that: if $p(x,y)=f(x,y)g(x,y)$, $f, g \in \bar{\mathbb{F}}[x,y]$ and $(a,b) \in \mathbb{F}^2$ is a zero of $p(a,b)=0$, then $p(a,b)=f(a,b)=g(a,b)=0$. In other terms if $p$ has a zero in $\mathbb{F}^2$, then this is an intersection point of the curves $f=0$ and $g=0$. Let's take for example $p=x^2+y^2$, which is irreducible in $\mathbb{R}[x,y]$. Then $x^2+y^2=(x+iy)(x-iy)$ and the only real root of $p$ is (0,0) which is the intersection point of $(x+iy)$ and $(x-iy)$.
Any other approach to the problem is obviously welcomed. Anyway this problem is intended for undergraduates with a good knowledge of algebra and galois theory, so it shouldn't involve any heavy machinery such as heavy algebraic geometry.
 A: Suppose that $p(x,y)$ is reducible in $\bar{\mathbb{F}}[x,y]$ and let  $g(x,y)\in \bar{\mathbb{F}}[x,y]$ be one of its irreducible factors.
Let $g=g_1,g_2,..., g_n (n\geq 2)$ be the conjugates of $g$ under the Galois group $G=\operatorname {Gal(\bar{\mathbb{F}}/\mathbb F)}$.
Since $g$ divides $p$, so do all the $g_i$'s: we have $$p=g_1\cdot g_2\cdot \ldots\cdot g_n\cdot \ldots\in \bar{\mathbb{F}}[x,y]$$
Now if $(a,b)\in \mathbb F^2$ is a zero of $g$, it will certainly be a zero of all the $g_i$'s.
In particular it will belong to the intersection $V(g_1)\cap V(g_2)\subset \bar{\mathbb{F}}^2$ of the curves $g_1=0$ and $g_2=0$.
But that intersection is finite, like all intersections of two different irreducible algebraic curves: Fulton  1.6,Proposition 2, page 9.    
Conclusion
The curve $p=0$ has indeed only finitely many zeros over $\mathbb F$. 
Edit
Why does $g$ have only many conjugates?
Because each coefficient of  $g$ has only finitely many conjugates:
Indeed given an element $a\in \bar {\mathbb F}$, let $f(X)\in \mathbb F[X]$ be its minimal polynomial.
Every conjugate $\gamma(a)\in \bar{\mathbb F}\; (\gamma\in G=\operatorname {Gal(\bar{\mathbb{F}}/\mathbb F)})$ is a zero of $f(X)$ too, so that the number of such conjugates is finite, bounded by the degree of $f(X)$ .
(This might appear a bit surprising, since $G$ itself will be infinite in general.)  
New Edit
The hypothesis "$\mathbb F$ has characteristic zero" ensures that the extension $\mathbb F\subset \bar{\mathbb F}$ is Galois.
As a consequence an element of $\bar F$ fixed by the Galois group $G=\operatorname {Gal(\bar{\mathbb{F}}/\mathbb F)}$ must belong to $\mathbb F$ and similarly for polynomials.
As a consequence when I wrote $p=g_1\cdot g_2\cdot \ldots\cdot g_n\cdot \ldots\in \bar{\mathbb{F}}[x,y]$, which was enough to conclude, we had actually $p=g_1\cdot g_2 \cdot \ldots \cdot g_n\in \bar{\mathbb{F}}[x,y]$ : the ellipsis "$\cdot \ldots$" is in fact non-existent.
Indeed the product $g_1\cdot g_2\cdot \ldots\cdot g_n$ is invariant under $G$ and thus has coefficients in $\mathbb F$ by what I just wrote.
Since it divides the irreducible polynomial $p \in \mathbb F[x,y]$, it must be equal to $p$. 
