I.e., why, when we talk about an algebraic curve $f(x, y) = 0$, do we assume that $f(x,y)$ is a polynomial?
Surely the analysis of a curve like $|\cos(z)|^2 = 1$ is not beyond the reach of mathematics -- let alone modern mathematics. But results like Bézout's Theorem require the curves in question to be polynomial. Is it that the extension to general analytic functions is too hard?
Or, on the other hand, is it too easy (e.g, do results for more general $f(x,y)$ follow easily from results for polynomial $f(x,y)$)?
You can study any equation you want. There's tons of mathematics for studying all kinds of equations. Some tools work better for this bunch of equations, other tools work better for that other bunch of equations.
It just so happens that there is an incredibly rich and beautiful set of tools for studying solution sets of polynomial equations, and which (for the most part) apply only to solution sets of polynomial equations and not other kinds of equations. This collection of tools is known as algebraic geometry.
As for your title question, what shall we call the solution sets of equations defined by polynomials? As the study of polynomial equations has evolved over the last two millenia or so, a terminology has emerged: nowadays their solution sets are known as algebraic varieties, and in the case of $f(x,y)=0$ as algebraic curves.
So, the real and literal answer to your title questions is: That's the definition of algebraic curves.