Let $K := \mathbb{C}(x,y : y^2 = x^6-1)$ (by this I mean the fraction field of $\mathbb{C}[x,y]/(y^2-(x^6-1))$) be the function field of your curve. "Points" on the smooth (=normalized) projective curve are the same thing as discrete valuations on $K$. You are looking for points where $x$ and $y$ are both "infinite" in the sense that they have negative valuation: but then $y^2 = x^6 - 1$ has valuation $2v(y) = 6v(x)$ so we have $v(y) = 3v(x)$. Let us assume for the moment that $v(x) = -1$, i.e., $1/x$ is a uniformizer (=element of valuation $1$), which of course implies that $v(y) = -3$.
To actually find and describe such valuations, I think the best practical way is to work with power series (this is what Newton would have done!).
Suppose you find two nonconstant elements $x,y$ of $\mathbb{C}(\!(t)\!)$ (the field of Laurent series) such that $y^2 = x^6 - 1$: they define a point of the (affine) curve in that field, and by the valuative criterion of properness and separatedness they extend to a unique point of the projective closure in $\mathbb{C}[\![t]\!]$, whose value at $t=0$ (now it makes sense) defines a $\mathbb{C}$-point of the projective curve, but in fact, one of any projective completion of the original affine curve, in particular, the smooth one. If you prefer to work with function fields, a different way of saying the same thing is that $x$ and $y$ define a field embedding $K \subseteq \mathbb{C}(\!(t)\!)$ for which the valuation on $\mathbb{C}(\!(t)\!)$ restricts to a valuation on $K$, i.e., gives a point on the smooth projective curve as mentioned above.
Now take a look at the series
$$y = t^{-3} - \frac{1}{2}t^3 - \frac{1}{8}t^9 + \cdots$$
(obtained by taking the square root of $1-t^6$ which starts with $1 - \frac{1}{2}t^6 - \frac{1}{8}t^{12} + \cdots$, and multiplying by $t^{-3}$), together with
$$x = t^{-1}$$
Clearly this defines a point $P_1$ as mentioned above. We find this sort of thing by starting with $x = t^{-1}$ since we wanted $1/x$ to be a uniformizer (essentially, we are looking at the completion of $K$ with respect to this uniformizer/valuation) and then we try to solve for $y$. But there is also a different solution, namely the other square root (the negative of the above)
$$-t^{-3} + \frac{1}{2}t^3 + \frac{1}{8}t^9 + \cdots$$
giving us another point $P_2$ on the normalizer, both above the (singular) point $P$ with coordinates $(X:Y:Z)=(0:1:0)$ of the singular projective curve (clearly $P_2$ is a different point from $P_1$ because the valuation of $y - x^3$ is going to differ between the two).
This is about as good a description of the two points as one could hope for, namely a power series "parametrization" of the two branches of the curve around the singular point $P$. There is then the matter of making sure that we have found all possible points above the singular point $P$ we were trying to resolve: this can be done by counting multiplicities: clearly $x$ defines a map to $\mathbb{P}^1$ that is of degree $2$ since $K$ is of degree $2$ over $\mathbb{C}(x)$, and we found two points $P_1,P_2$ over $\infty \in \mathbb{P}^1$, so we found them all (and neither is ramified over $\infty$). (I wish I had a cleaner, less ad hoc, argument to explain why we must have $v(x) = -1$, though.)
This is slightly more efficient than considering a sequence of blowups: here, if we work in the local coordinate chart given by $\tilde z := Z/Y = 1/y$ and $\tilde x := X/Y = x/y$ (satisfying $\tilde x^6 = \tilde z^4 + \tilde z^6$, as you mention), we find that $\tilde z$ looks like $t^3$ and $\tilde x$ looks like $t^2$ for one of the points on normalization, and their opposites for the other. So around $P$ with $(\tilde z, \tilde x) = (0,0)$, both curve branches we found have a cusp with the same tangent line, and it takes at least two successive blowups to resolve them. As for the connection with the integral closure, it is going to be the set of elements of $K$ which have nonnegative valuation everywhere on the affine chart being considered (including the two points we found on the normalization): in the case of $\tilde z,\tilde x$, for example, we can form $\tilde x^3/\tilde z^2 = x^3/y$, which is not in $\mathbb{C}[\tilde z,\tilde x]$ (because it takes a different value at $P_1$ and $P_2$) but is in its integral closure (the places where it is infinite are those with $y=0$ which are not in the affine patch covered by $\tilde z,\tilde x$).
As you said, any smooth integral curve $Y$ (over an algebraically closed field) is a Zariski open subset of a projective curve $Z$. Let $\pi: X\to Z$ be the normalization. Then, $X$ is smooth (integrally closed=smooth in this situation, for curves). Also, $\pi^{-1}(Y)\to Y$ is an isomorphism, since $Y$ is smooth. Thus, $Y$ is an open set of $X$, which is smooth.
For any integral curve (say $X$), the closed sets are, 1) $X$; 2) $\emptyset$; 3) finite set of points. It should be clear now why $Y$ is the complement of finite set of points in $X$.
The final point, there are several arguments, may be the one I describe is not the simplest. If $Y\subset X, X'$ both $X, X'$ smooth projective, then using identity map on $Y$, we get a rational map $f:X\to X'$. But birational maps from smooth (projective) curve to a smooth projective curve is in fact an isomorphism. This says, the cardinality of $X-Y, X'-Y$ are the same. Also note that since $X,X'$ are isomorphic, their genus are same, giving you uniqueness of $(g,r)$.
Best Answer
It's not even true in that specific example that you're only adding one point at infinity: the part of that curve at infinity is given by $V(3X^3+4Y^3+5Z^3)\cap V(Z)=V(3X^3+4Y^3,Z)$ which is three points, $[1:\sqrt[3]{\frac34}:0]$, $[1:\omega\sqrt[3]{\frac34}:0]$, and $[1:\omega^2\sqrt[3]{\frac34}:0]$ where $\omega$ is a primitive cube root of unity.
The general situation is that for any smooth projective curve, any nonempty proper open subvariety is affine, so by starting with a projective curve and removing an arbitrary finite number of points $n$ we may find a smooth affine curve so that the (smooth) projectivization needs $n$ additional points to be added to the curve.
Let me also warn you that it is frequently false that the projectivization of a smooth affine curve is a smooth projective curve: for instance, $y=x^3$ which is smooth projectivizes to $YZ^2=X^3$ which has a cusp at $[0:1:0]$.