I solved this problem, but I'm not sure my answer is correct as it seems very complex (compared to the polar equation). Did I make some mistake along the way or is it the right solution?
$$r=\theta$$ $$\sqrt{x^2+y^2}=\arctan\bigg(\frac{y}{x}\bigg) \tag{$r^2=x^2+y^2 \to r=\sqrt{x^2+y^2}$}$$ $$x^2+y^2 = \bigg(\arctan\bigg(\frac{y}{x}\bigg)\bigg)^2$$
On
$\arctan$ is a tricky function. Remember, $\arctan(\tan(\theta))$ does not always equal $\theta$. Sometimes $$\theta=\arctan(\tan(\theta))+k\pi$$
The Cartesian version of your equation is $$\sqrt{x^2+y^2}=\arctan\left(\frac{y}{x}\right)+k\pi$$ over all integer values of $k$. Or more concisely:$$\tan\sqrt{x^2+y^2}=\frac{y}{x}$$ But even this equation doesn't capture the points on the spiral that are on the $y$-axis, where $x=0$ and $y=\pi/2+k\pi$ cause neither of the two sides to be defined. This can be "fixed" by declaring that any pairs$(x,y)$ causing both sides to be undefined should also be included in the curve. All such points are of the form $(0,\pi/2+k\pi)$.
$\quad r^2 = \theta^2\quad$ expressed as a Cartesian equation is
$$x^2+y^2 = \bigg(\arctan\left(\frac{y}{x}\right)\bigg)^2$$
But better yet, for $r = \theta$, use: $$\sqrt{x^2+y^2} = \arctan\left(\frac{y}{x}\right) + k\pi\tag{see alex's explanation}$$
ADDED: This is the graph I get (from Wolfram) for the function: using a polar plot.