Hilbert's Axioms of incidence for lines are
I1. For any two distinct point A, B there exists a unique line $l$ containing A, B.
I2. Every line contains at least two points.
I3. There exist three noncollinear points (that is, three points not all contained in a single line).
We see that the set $\mathbb R \times \mathbb R$ as points and the set of points that satisfy equations of the form $ax+by+c=0$ as lines are a model for the above three axioms.
My question is, is there a different definition of a line in the cartesian coordinates (that is not algebraically equivalent to this definition) that still satisfy the above three axioms? Or is the form $ax+by+c=0$ fundamentally the only form that can be considered a line? If so, why?
There are many ways to model points as elements in $\mathbb R\times\mathbb R$ without necessarily modelling lines as solutions of linear equations.
Here is a generic way of doing so. Create a bijection $f:\mathbb R\times\mathbb R\to\mathbb R\times\mathbb R$. Any bijection, no particular properties needed. Now, that bijection is going to map "ordinary" points into "ordinary" points, but it may not map "ordinary" lines into "ordinary" lines. Take an example: $f(x,y)=(x,\sqrt[3]{y})$. An ordinary line $ax+by=c$ is mapped by $f$ into the following set of points: $ax+by^3=c$ - those are certainly not straight lines but cubic curves.
Still, by using the fact that $f$ is a bijection and that it maps "incidence" into "incidence", you can conclude that the new model, where points are elements of $\mathbb R\times\mathbb R$ but the lines are curves with equations $ax+by^3=c$ satisfies any incidence axioms that the original model (with "ordinary" lines satisfies), and vice versa.
NB: In other words, you have not only found a non-standard model of Hilbert's axioms, you've found a model isomorphic to the standard one (the isomorphism being the bijection $f$). The question whether there are other, non-isomorphic, models on $\mathbb R\times\mathbb R$ (i.e. is every model on $\mathbb R\times\mathbb R$ of this type with a suitable choice of $f$) would be a separate question, I don't know the answer to it on the top of my head.