Is the set $S=\{(x,y)\in\Bbb R^2\mid e^{x^2+y^2}=2+x^2+y^2\}$ bounded?

Is the set $S=\{(x,y)\in\Bbb R^2\mid e^{x^2+y^2}=2+x^2+y^2\}$ bounded?

My thoughts:

Let's consider the function $g:\Bbb R\to\Bbb R, g(x)=e^x-x-2.$ Checking its derivative, we see $g$ is strictly increasing on $[0,+\infty)$. Furthermore, $g(0)=-1$ and $\lim\limits_{x\to+\infty}g(x)=+\infty,$ which means $g$ changes signs and has a unique root, call it $r_0$ in $[0,+\infty).$ My conclusion is that $S$ is a circle with a radius $r_0$ centered at the origin, hence bounded.

I'm sceptical about my deduction and answer. Is there any room for improvement?

What you did are all correct. But if all you want is the boundedness, then $\lim_{x\rightarrow\infty} g(x)=+\infty$ is all you need. Indeed, this implies that there exists $r>0$, such that for all $x>r$, $g(x)>0$. Hence any point $(x,y)$ on $S$ must satisfy $x^2+y^2\le r$, thus $S$ is bounded.

This is helpful for more complicated problems. For example, let $S=\{(x,y)\in\mathbb R^2 : e^{x^2+y^2} = 2 + x^2 + 10000y\}$ which is not any familiar curve. We still consider $g$, and use the fact that when both $y$ and $x^2+y^2$ are sufficiently large, $e^{x^2+y^2}>2+x^2+y^2>2+x^2+10000y$. Hence for $(x,y)$ on $S$, $x^2+y^2$ (and $y$) has to be small.

It's often beneficial to think roughly than precisely in analysis.

Looks good other than the crime of not converting $x^2 + y^2$ to polar coordinates. Substituting $x^2 + y^2$ for x isn't something I'd do.

Let $x=r\cos(\theta), y=r\sin(\theta)$

$\implies x^2 + y^2 = r^2\cos^2(\theta) + r^2\sin^2(\theta) = r^2(\cos^2(\theta) + \sin^2(\theta))=r^2$.

Then $S=\{(x,y)\in\Bbb R^2\mid e^{x^2+y^2}=2+x^2+y^2\}$

$=\{(r, \theta) \in\Bbb R^2 \mid e^{r^2}=2+r^2\}$.

Let $g(r)=e^{r^2}-2-r^2$

$\implies g'( r) = e^{r^2}\cdot 2r - 2r = 2r(e^{r^2} -1)$.

Since $\forall r\ge 0, g'(r)\ge0$

$\implies S$ is bounded.

But I'm open to others' opinions.