What I would like to do is find the solutions of $p_1(x,y) = x^2 + y^2 - 1 = 0$, after passing to the projective plane.
Step 1, of course, is to homogenize $p_1$. Then we have $\tilde{p_1}(x,y,z) = x^2 + y^2 - z^2$.
Now I have proven two things already that simplify my work. I have proved that if a point $(x,y,z) \in \mathbb{C}^3 - (0,0,0)$ satisfies the homogeneous polynomial, then $(x : y : z)$ (all the points in the equivalence class of $(x,y,z)$) satisfy the homogeneous polynomial. It has also been shown that if $(x,y)$ is in the $V(p_1)$, then $(x,y,1)$ is in $V(\tilde{p_1})$. From this I believe we can conclude that the only points left to find, are the points at infinity?
(this is the part where I am confused)
So to find the points at infinity in $V({\tilde{p_1}})$ do we just set $z = 0$ and solve
$$x^2 + y^2 = 0?$$
This would yield $|x| = i|y|$? This makes sense looking at the equation but it feels weird expressing this as a point of $\mathbb{P}^2_{\mathbb{C}}$. How could I write it in the form $(a : b : 0 ) \in \mathbb{P}^2_{\mathbb{C}}$? How can I visualize the line that the point represents? Any geometric intuition?
You are more or less correct. The line at infinity is given by $z=0$ and so
$$x^2 + y^2 = 0 \Rightarrow x = \pm iy $$
So there are actually two more points $[i,1,0], [-i,1,0]$ at infinity.
For geometric intuition, note that
$$x^2 +y^2 =1 \Rightarrow (x+iy)(x-iy) =1.$$
So under the change of coordinate $z= x+iy, w = x-iy$, the equation becomes
$$ zw=1$$
and one can think of this as a graph $\{(z, 1/z) :z\in \mathbb C\setminus \{0\}\}$. Thus the equation represents something like $\mathbb C\setminus \{0\}.$ Since the object in $\mathbb{CP}^2$ is like compactifying the original object in $\mathbb C^2$, so in this situation the two points added are "$z=0$ and $z = \infty$ (which is $w=0$)" The first one is $x-iy = 0$ and the second one is $x+iy = 0$.
Note also that the closure in $\mathbb{CP}^2$ actually looks like a $2$-sphere (or $\mathbb{CP}^1$).