Let’s go back to the original polynomial,
$$x^3+x^2y-x^2+2xy+y^2-2x-2y\;.\tag{1}$$
That $2xy$ looks a lot like the middle term of $(x+y)^2$, and the $-2x-2y$ can certainly be written nicely in terms of $x+y$, so let’s try something along those lines. $(x+y)^2=x^2+2xy+y^2$, and we have the $2xy+y^2$, but instead of $x^2$, we have $-x^2$. In other words, we have $(x+y)^2-2x^2$. That’s not entirely promising, but let’s see where it goes. We can rewrite $(1)$ as
$$\begin{align*}
(x+y)^2&-\;2x^2-2(x+y)+x^3+x^2y\\\\
&=(x+y)^2-2(x+y)+x^2(x+y-2)\\\\
&=(x+y-2)(x+y)+x^2(x+y-2)\\\\
&=(x+y-2)(x^2+x+y)\;.
\end{align*}$$
Added: Alternatively, following the hint, interchange the rôles of $x$ and $y$, rewriting $(1)$ as $$y^2+(x^2+2x-2)y+(x^3-x^2-2x)\;.\tag{2}$$
Remember that linear factors of a polynomial correspond to zeroes. Set $(2)$ equal to $0$ and use the quadratic formula to solve for $y$:
$$\begin{align*}
&(x^2+2x-2)^2-4(x^3-x^2-2x)\\
&\qquad=x^4+4x^3-8x+4-4x^3+4x^2+8x\\
&\qquad=x^4+4x^2+4\\
&\qquad=(x^2+2)^2\;,
\end{align*}$$
so $$y=\frac{-x^2-2x+2\pm(x^2+2)}2\;,$$
and $y=-x^2-x$ or $y=-x+2$. Thus, $(2)$ factors as $$\Big(y-(-x^2-x)\Big)\Big(y-(-x+2)\Big)=(y+x^2+x)(y+x-2)\;.$$
Here's an attempt to put the problem in a somewhat broader setting, while avoiding Galois Theory.
Let $P$ be a polynomial with integer coefficients, irreducible over the rationals.
Let $P=pq=rs$ where $p,q$ have coefficients in $K$ and are irreducible over $K$, $r,s$ have coefficients in $L$ and are irreducible over $L$, and $K\ne L$.
Renaming the polynomials, if necessary, we may assume there exist $\alpha,\beta$ such that $p(\alpha)=r(\alpha)=0$, $p(\beta)=0\ne r(\beta)$. Then there are polynomials $a,b$ with coefficients in the field $M$ generated by $K$ and $L$ such that $ap+br=0$, with the degree of $a$ less than that of $r$, and the degree of $b$ less than that of $p$. Now if you evaluate at $\beta$ you get $b(\beta)=0$ which gives you further information on the degree of $b$. I think in the setting of the original question, this is enough to prove impossibility; I think that in general, it's enough to give strong conditions on what degeres are possible.
Best Answer
The roots of any polynomial over $\mathbb{F}_3$ lie in some finite extension field $\mathbb{F}_{3^n}$.
The roots of $x^4 + 1$ are all primitive eighth roots of unity. Therefore, we know they live in any extension field such that $8 \mid 3^n - 1$.
The smallest such field is the one with $n=2$. However, this is merely a degree 2 extension! The minimal polynomial of an eighth root of unity in characteristic 3, then, is a quadratic polynomial. The irreducible factorization of $x^4 + 1$ over $\mathbb{F}_2$, therefore, must be a product of two quadratic polynomials.
We can say more: if $\zeta$ is a primitive eighth root of unity, then $\zeta$ and $\zeta^3$ are roots of one factor, and $\zeta^5$ and $\zeta^{15} = \zeta^7$ are roots of the other factor. We have a factorization
$$ x^4 + 1 = (x^2 - (\zeta+\zeta^3) x + \zeta^4) (x^2 - (\zeta^5 + \zeta^7) x + \zeta^4) $$
If you construct a realization of the field $\mathbb{F}_9$, you could do the arithmetic to simplify the coefficients to elements of $\mathbb{F}_3$.
But we can cheat: from ordinary arithmetic in characteristic 0, we know the standard eighth roots of unity
and so the factorization should simplify to
$$ x^4 + 1 = (x^2 - \frac{\mathbf{i}\sqrt{2}}{2} x + (-1))(x^2 - \frac{-\mathbf{i}\sqrt{2}}{2} + (-1))$$
In characteristic 3, $(\mathbf{i} \sqrt{2})^2 = -2 = 1$, so if we pick $1$ as the product of square roots, we should have
$$ x^4 + 1 = (x^2 - \frac{1}{2} x + (-1))(x^2 - \frac{-1}{2} + (-1)) = (x^2 - 2x - 1)(x^2 + 2x - 1)$$
and if we didn't believe that this "cheat" was actually a valid derivation, we can verify this factorization is correct directly.