What is the cardinality of the set $ \{ z \in \mathbb C : e^{z} =z \}$?
In other words, what is the cardinality of the zeros of the function $ f(z) = e^{z} - z $? Is it finite?
Does Rouche's theorem work here?
Moreover, what about the cardinality of the set $ \{ z \in \mathbb C : | e^{z}|= |z| \} $? Is this helpful to conclude?
In general, an entire (nonzero) function $f$ can have finitely many zeroes iff $f(z)=P(z)e^{g(z)}$ for some possibly constant non-zero polynomial $P$ and an entire function $g$; this is easily seen by factoring out the zeroes with a polynomial and then noting that any entire function $f/P$ which doesn't vanish has a holomorphic logarithm $g$
In the case that $f$ is of finite order ($\log (\max_{|z|=R} |f(z)|) /R^q$ bounded for some finite positive $q$ (or equivalently $|f(z)| \le C_q e^{|z|^q}$), it is immediately seen that any $g$ as above must be a polynomial of degree at most $q$ and in particular if the order (the infimum of all $q$ as above) is not integral, we get a contradiction, so $f$ must have infinitely many zeroes and there is a nice theory that gives their density and more.
If however $q$ is integral as in our case $f(z)=e^z-z$ where $q=1$, it may happen that the function is indeed $P(z)e^{g(z)}$ so we need to eliminate that possibility; here since $q=1$ we must eliminate the possibility $g(z)=az$ since $g$ has at most degree $1$ and the constant ($e^b$ if $g(z)=az+b$) can be absorbed into $P$.
So assuming $e^z-z=P(z)e^{az}$ we first can take $z=\pm iR$ and get that the imaginary part of $a$ is zero since $e^{ia(\pm R)}$ grows like $e^{\pm (R \Im a)}$ but $e^{iR}-iR$ grows just like $R$; then by taking $z=R$ we get first $a=1$ so the exponential growth rates match and then $P$ constant $1$ as $1-Re^{-R} \to 1, R \to \infty$ hence $P(R) \to 1, R \to \infty$ but clearly $e^z-z \ne e^z$.
The proof above shows that for any non-constant polynomial $Q$, $e^z-Q(z)$ has infinitely many roots
As for the second question, note that if $|z|=R, \Re z=x, |x| \le R$ the equation $|e^z|=|z|$ becomes $e^x=R$ or $x=\log R$ and the only condition we need is $|\log R| \le R$ where if we get strict inequality we have two solutions for $z$, while if we have equality (there is an unique point in $(0,1)$ for which that happens) we have a unique $z$,so we have 2 solutions on each circle for $R >r_0$, one at $R=r_0$ and none for $0 \le R < r_0$ where $r_0$ is the unique solution (obviously in $(0,1)$) of the equation $R+\log R=0$