Am I applying MAX Modulus principle correctly here?

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$(1)$ Find all possible entire functions s. t. $$|f(z)|\leqslant2|z|+1.$$
$(2)$ Prove you have found all such functions.

According to my understanding of $MAX$ modulus principle, $|f(z)|$ doesn' t have a $\max$ in a domain $D$, if it is not constant in $D$, but if it is continuous on a closed, bounded region of $D$, then even if $f$ is not constant, it attains its max, on the boundary of the region.

So here, since $|f(z)|$ has a max for ALL z, by MMP, $f(z)$ must be constant.
but then I randomly tried other functions for example $f(z)=z^2$ also works for some $z$ ( well not for all $z$ ). I am not sure if I am correct and how to show I' ve found ALL such functions.

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$$f''(z)=\frac{2!}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^3}dw$$

Where, $\gamma : |w-z|=r $

(And above integral is independent of any $r>0$, as $f$ is an entire function.)

So we have,

$$\pi|f''(z)|\leq \int_{\gamma}\frac{|f(w)|}{|(w-z)^3|}|dw|\leq \int_0^{2\pi} \frac{2|w|+1}{r^3}(rdt)\leq \frac{2|z|+2r+1}{r^2}(2\pi) \tag{$\forall r>0$}$$ (Since, $ \text{sup}\{|w| \ | w\in \gamma\}=|z|+r$ )

Taking $r \to \infty$,we get

$f''(z)=0$

Hence , $f(z)=az+b$ for some $a,b\in \Bbb C$

Using similar argument with $f'(z)$ and $f(0)$ we can infer $|a|\leq 2$ and $|b|\leq 1.$... (Which are neccessary conditions.)

Now , $|f(z)|=|az+b|\leq 2|z|+1$

Hence these neccessary conditions are sufficient as well.