When does an irreducible polynomial stay irreducible as a power series? My question is inspired by this topic. 

Is there a way to decide if an irreducible polynomial in $\mathbb C[x,y]$ remains so in $\mathbb C[[x,y]]$? 

For example, the polynomial $p(x,y)=y^2-x^{2t}a(x)$, $a(0)\neq 0$, is irreducible when $a(x)$ is not a square, but $p(x,y)$ is reducible in $\mathbb C[[x,y]]$. 
 A: There is a factorization method that works for any power series $p(x,y)\in\mathbb C[[x,y]]$, due to Isaac Newton (rediscovered by Puisseux).  Here is how it works in this case. Notice that $p$ is a unit iff its absolute term is non-zero. For an arbitrary $p$ let us suppose
that neither $x$ nor $y$ divides $p$ (otherwise divide it; or if you ask about irreducibility of $p$, then it is not).
Observation: if $m\in\mathbb N$ is smallest such that $y^m$ is in $p$ with non-zero coefficient, then $p$ is a unit times a monic degree $m$ polynomial (in $y$) $p'\in\mathbb C[[x]][y]$ (the needed unit is easily found inductively and is unique). The factorization problem in $\mathbb C[[x,y]]$ is thus the same as the factorization problem for monic polynomials in $\mathbb C[[x]][y]$.
This problem has an algorithmic solution (well, modulo finding roots of complex polynomials). Namely, if $K$ is algebraically closed (for us $K=\mathbb C$) then the field of Puisseux series $K\{\{x\}\}$ (=Laurent series in $x^{1/m}$ with finitely many negative powers, union over all $m$'s) is also algebraically closed, and there is an algorithmic proof of this fact. Below is a partial description of the algorithm and of the factorization of $p'$ it gives. 
For every $x^my^n$ with non-zero coefficient in $p$ (or in $p'$) draw the point $(m,n)$ to the plane, and also draw all the points $(k,l)$ with $k\geq m$, $l\geq n$. The convex hull of the resulting set is the Newton polygon $\Delta_p=\Delta_{p'}$ of $p$. It is an "infinite polygon", with two infinite edges along the axes, and a finite number of finite edges.
As an elementary observation, $\Delta_{pq}=\Delta_p+\Delta_q$. That already shows that if $\Delta_p$ has a single finite edge, and no integer point on that edge (except for its ends), then $p$ is irreducible in $\mathbb C[[x,y]]$.
If $\Delta_p$ has $k$ finite edges then $p'$ factorizes to (at least) $k$ polynomials, so for $k\geq 2$ it is not irreducible.
In the remaining case one needs to do an actual calculation. Suppose that $\Delta_p=\Delta_{p'}$ has a single finite edge, which contains (inside) $k-1$ integer points (i.e. is composed of $k$ segments with integer endpoints). Certainly $k|m$ ($m$ is the degree of $p'$). One needs to apply Newton (or any other) method to solve $p'(x,y)$ for $y$ in terms of Puisseux series in $x$, just far enough to find the least common denominator of the exponents of $x$ in $y(x)$. If it is $m$ then $p$ is irreducible, if not, it is reducible. 
