Let's assume elliptic curve $E$ over $\mathbb{R}$:
$y^2 = x^3 + x + 1$
How to convert this equation to homogeneous coordinates?
My notes say it's $zy^2=x^3+xz^2+z^3$. Unfortunately, I have no idea where that came from and whether it's correct or not.
No, you don't just "add $z$'s" until the equation is homogeneous, and no, you don't substitute $z=1$, as the comments above may suggest. This works, but that's not the reason, and not what is formally happening.
Let's work backwards. Let's start with a curve given in projective coordinates, let's say $$C: f(X,Y,Z)=ZY^2 - X^3 - XZ^2-Z^3=0,$$ i.e., $$C=\{[x_0,y_0,z_0]\in\mathbb{P}^2 : f(x_0,y_0,z_0)=0\}.$$ There is a map from $C$ to an "affine patch" $E$ of $C$ given by such that $$\phi([X,Y,Z]) = (X/Z,Y/Z)$$ which is well-defined for all points on $E$ with $Z\neq 0$. In other words, $\phi:C\setminus\{[0,1,0]\} \to E$. This is called de-homogenizing the equation of $C$ (note that there are other affine patches we could take). We also note that if $f(X,Y,Z)=0$, and $Z\neq 0$, then dividing through by $Z^3$ we obtain $$\frac{Y^2}{Z^2} - \frac{X^3}{Z^3} -\frac{X}{Z}-1=0$$ or, in other words, $(X/Z,Y/Z)$ satisfy the equation $y^2=x^3+x+1$, so the affine patch $E$ is the curve defined by this affine equation.
Conversely, if we begin with an affine patch of a curve $E:y^2=x^3+x+1$ we find the projective model by inverting $\phi$, i.e., set $$C: (Y/Z)^2 =(X/Z)^3+(X/Z)+1,$$ and multiply through by $Z^3$ to obtain $$C : ZY^2 = X^3 +XZ^2+Z^3.$$
In general, the homogeneous equation of a curve $E:f(x,y)=0$, where the highest degree of a monomial of $f$ is $d$, is given by $$C:Z^df(X/Z,Y/Z)=0.$$