Question as in the title. Also, if a $2$-dimensional parametric equation does not have a Cartesian form, how would one go about proving this?
Two particular parametric equations I believe do not have a Cartesian form. (Or at least, I was unable to find the Cartesian form.)
First set of equations: (The square roots are difficult to eliminate.)
$x = \sqrt{60t} - t$
$y = (\sqrt{60t} - t)\left(\frac{60}{\sqrt{60t}}-1\right)$
Second set of equations: (The term $f(t)$ is difficult to eliminate.)
Let $f$ be some differentiable relation and let $r \in \mathbb{R}$. Define
$x = t - r\sin(\arctan(f'(t)))$
$y = f(t) + r\cos(\arctan(f'(t)))$
One interesting thing to note about the second set of parametric equations is that the set of points $(x,y)$ is an exact duplicate of $f(t)$, but $r$ units away along the normal through $(t,f(t))$.
I believe that you can always make one variable in an equation of two variables the "subject" of the equation. Maybe not in terms of elementary functions in very specific cases, but I'm pretty sure it's always possible. (And then substituting what you found for $t$ into the other equation is trivial.)
So, for instance, with your first set:
$$\sqrt{60t} - t = x$$ $$\sqrt{60t} = x + t$$ $$60t = (t+x)^2$$ $$60 t-t^2-2 t x-x^2 = 0$$ Solve this quadratic in $t$ to get $$t = 30 - x \pm \sqrt{900-60x}$$
Plugging in is the easy part. Replace $t$ with $30 - x \pm \sqrt{900-60x}$ to get
$$y = \sqrt{60} \, \Bigg(-1 + \frac{\sqrt{60}}{\sqrt{30 \pm \sqrt{900 - 60 x} - x}} \Bigg) \sqrt{30 \pm \sqrt{900 - 60 x} - x}$$ which looks like