Delta functions are just shorthand notation to restricting integration to a lower dimensional region (in this case, a circle). Delta function only makes real sense inside an integral, and has its own rules for changing variables.
In this case, just change variables to polar coordinates (not forgetting the Jacobian):
$$\int_{-\infty}^\infty\int_{-\infty}^\infty \delta(x^2+y^2-R^2)dx\,dy=
\int_0^{2\pi}\int_0^\infty \delta(r^2-R^2)r\,dr\,d\phi$$
The $\phi$ integral is trivial:
$$\color{red}{=2\pi\int_0^\infty \delta(r^2-R^2)r\, dr }$$
Consider now the defining property of the delta function that says it just evaluates the function at its peak position:
$$\int f(x)\delta(x-x_0)dx=f(x_0)$$
The other important part is the variable substitution rule:
$$\delta(g(x))=|g'(x_0)|^{-1}\delta(x-x_0)$$
where $x_0$ is the zero of $g(x)$.
In our case, $g(r)=r^2-R^2=(r-R)(r+R)$ has a zero at $r=R$ (the negative one is outside the integration range). We have $g'(R)=2R$. So, the above integral in $\color{red}{\text{red}}$ can be transformed into
$$=2\pi\int_0^\infty \frac{1}{2R}\delta(r-R)r\, dr=\pi$$
where we took into account that this delta function just evaluates the function $\frac{r}{2R}$ at $r=R$.
Notice that it is extremely important that the argument of the delta function is not $\sqrt{x^2+y^2}-R=r-R$ but $x^2+y^2-R^2=r^2-R^2$.
Another way of looking at this is to start from the red integral again, but change variables to $r^2=u$.
$$2\pi\int_0^\infty \delta(r^2-R^2)r\, dr=
2\pi\int_0^\infty \delta(u-R^2)\frac{du}{2}=\pi
$$
where the delta function now just evaluated function $\frac{1}{2}$ at $u=R^2$ which doesn't matter because the function was just a constant.
The moral of the story: it's not just important where the zero of the delta function's argument is, you have to take into account the functional form (especially the derivative). Stretched delta isn't the same as delta even if the peak is at the same place. If you imagine it as a narrow and tall Gaussian peak, you can see that stretching the function by changing variables changes its area.
You could also write your delta function as $$\delta(r^2-R^2)=\delta((r-R)(r+R))\equiv \delta(2R(r-R))\neq \delta(r-R)$$ where I took into account that the pre-factor $r+R$ is only important where $r=R$, so I put that in.