So if $f(z)=az+b$, is an entire injective map from $\Bbb C$ to $\Bbb C$, for both a and b are complex numbers and a is not equal to 0.
I've proved that if f is an injective entire function, it cannot have an essential singularity at infinity, but then how to show that $f(z)$ has to be a polynomial? And why $f(z)$ has to satisfy that $f(z)=az+b$?
First since $f$ is an injective entire function, $\lim\limits_{z \to \infty} f(z) = \infty$. (This is because if $f$ is bounded entire, then $f$ would be constant.)
Next suppose $f$ has Taylor series $$ f(z)=\sum\limits_{n=0}^{\infty}a_nz^n \quad\text{and }\quad g(z) =f(1/z) =\sum\limits_{n=0}^{\infty}\frac{a_n}{z^n} $$ If $f$ is not polynomial, then $0$ is an essential singularity of $g$ ($\infty$ is an essential singularity of $f$). By Casorati-Weierstrass theorem, for any $A\in \Bbb{C}$, there is a sequence $z_n\to0$ such that $\lim\limits_{n\to\infty}g(z_n)=A$, i.e. there is $z_n'=1/z_n\to\infty$ such that $\lim\limits_{n\to\infty}f(z_n')=A$, contradicting $\lim\limits_{n\to\infty}f(z)=\infty$.
Finally any polynomial with degree more than one has more than one root and thus can't be injective.