I don't really know what you mean when you ask "Why is that a curve?". The question can be taken to mean either "I don't know what an algebraic curve is", or "Since there are three unknowns and two equations, it looks more like a surface to me". I am going to assume the latter.
So recall that a curve is most often defined either by a set of homogeneous equations in $n+1$ variables, in which case their solutions form a well defined subvariety of the projective space $\mathbf P^n$, or a set of arbitrary equations in $n$ variables, in which case their solutions are taken inside of affine space $\mathbf A^n$. We can go from one to the other by setting one of the variables equal to one and homogenizing respectively (but note that this process is not bijective).
So note first that our equation is not homogeneous. Based on the previous paragraph it would make most sense to interpret it as a subvariety of $\mathbf A^3$, but this is clearly not what they mean since that would give us an algebraic surface and not an algebraic curve.
Here are two ways to make sense of this:
Set $z=1$, for example. Then we get the equation $2y^2 = x^4 - 17$ in two variables, and whenever we have a solution to this equation we also have a solution to the original one. This equation defines a plane curve in $\mathbf A^2$. By embedding $\mathbf A^2$ in a projective space and taking the closure we get a projective curve, and by desingularizing it we also get a smooth projective curve.
Give the variable $y$ degree $2$, so that the equation is homogeneous with respect to a shifted degree. Then we can talk about the set of solutions inside of a weighted projective space. (Over $\mathbf C$ this can be defined as $\mathbf C^3 / \mathbf C^\ast$ where $\mathbf C^\ast$ acts by $\lambda \cdot (x_0,x_1,x_2) = (\lambda^2 x_0, \lambda x_1, \lambda x_2)$.)
Your second question - how do we see that this curve has genus one - can be determined in several ways. I am not sure what is the simplest way but here is one, using model (1) above and calculations in affine coordinates. Your curve has a degree 2 map to $\mathbf A^1$ given by projection to the $x$-coordinate, and this map will have branch points wherever $x^4-17 = 0$. The map can be extended to a map from the smooth projective version of the curve to $\mathbf P^1$ (this is a general fact about smooth curves, but you can also prove this by computing in explicit coordinates, see the Wikipedia article on hyperelliptic curves). Using the Riemann-Hurwitz formula one sees that:
a map of degree two must have an even number of branch points, so the point $x = \infty$ can not be a branch point;
a curve which has a degree two map to $\mathbf P^1$ with four branch points has genus one.
Oh, and please tell me which parts of this need elaborating.