Prove $X^2+Y^2-1$ is irreducible using geometrical tools.

1. A line in $k^2$ has the form $aX+bY+c=0$ where $a$ and $b$ cannot both be zero. We may assume without loss of generality that $a\ne 0$, so that we have $X=-a^{-1}bY-a^{-1}c$. It follows that for any $\alpha\in k$, the point $$(-a^{-1}b\alpha-a^{-1}c,\alpha)$$ lies on the line. It follows that the number of points on any line is in bijective correspondence with the field $k$. So in order for a line to have at least three points, we should assume that $k$ is not the field of $2$ elements, which is not a very strict assumption.
2. If $X^2+Y^2-1=F(X,Y)\cdot G(X,Y)=0$, then notice by the zero product property that any point on the circle must satisfy at least one of the linear polynomials $F(X,Y)$ and $G(X,Y)$. It is also true that any point satisfying the polynomials $F$ and $G$ will lie on the circle. It follows that the circle must be the union of two lines (a contradiction).

The "proof" is absolute nonsense:

If you replace the word "circle" by "degenerate conic" (= union of two lines, non necessarily distinct) the "argument" still applies, which proves that it is false, since degenerate conics do exist!
And actually the circle is reducible over an algebraically closed field of characteristic $2$, and such a field is necessarily infinite, so that lines have infinitely many points.
This confirms that the purported proof is no proof at all.

The mistake is the statement "a line and a circle have at most two points in common": this is false if the circle contains a line and since what we want to prove is that the circle contains no line we may not assume it in the proof: the argument is circular (pun intended!)

If the field $k$ is the finite field of 2 elements then this argument cannot work - there are lines in $\mathbb{F}_2[x,y]$ whose locus includes only two points, e.g. $x+y=1$, and indeed $x^2 + y^2 - 1 = (x+y+1)^2$ over $\mathbb{F}_2$. So, let's assume $\text{char } k$ is at least 3.

In that case, a line of the form $ax+by = c$ has at least 3 solutions $(x_0,y_0)$. Since our field $k$ has at least 3 different elements, we can substitute each of these in for x or y (a or b could be 0) and solve for the remaining variable. This gives us three solutions to the equation $ax+by=c$.

If $x^2 + y^2 - 1 = F(x,y) \cdot G(x,y)$, then any solution $(x_0,y_0)$ to the equation $x^2 + y^2 - 1 = 0$ must also then satisfy $F(x,y) = 0$ or $G(x,y) = 0$. This means that the circle $x^2 + y^2 - 1 = 0$ is actually a union of two lines.