We will prove a slightly more genaral fact. The proof is based on this answer
Theorem. Let $f\in\mathcal{O}(\mathbb{C})$ and for all $z\in\mathbb{C}$ we have $|f(z)|\leq\varphi(|z|)$. Assume that $$ \lim\limits_{R\to+\infty}\frac{\varphi(R)}{R^{p+1}}=0 $$ then $f$ is a polynimial with $\deg (f)\leq p$.
Proof. Since $f\in\mathcal{O}(\mathbb{C})$, then $$ f(z)=\sum\limits_{n=0}^\infty c_n z^n $$ for all $z\in \mathbb{C}$. Moreover, for all $R>0$ we have integral representation for coefficients $$ c_n=\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz $$ Then, we get an estiamtion $$ |c_n|\leq \oint\limits_{\partial B(0,R)}\frac{|f(z)|}{|z|^{n+1}}|dz|\leq \oint\limits_{\partial B(0,R)}\frac{\varphi(|z|)}{|z|^{n+1}}|dz|= \frac{2\pi R\varphi(R)}{R^{n+1}}= \frac{2\pi \varphi(R)}{R^{n}} $$ Hence for $n>p$ we obtain $$ |c_n|\leq\lim\limits_{R\to+\infty}\frac{2\pi\varphi(R)}{R^n}= 2\pi\lim\limits_{R\to+\infty}\frac{1}{R^{n-p-1}}\lim\limits_{R\to+\infty}\frac{ R\varphi(R)}{R^{p+1}}=0 $$ which implies $c_n=0$ for $n>p$. Finally we get $$ f(z)=\sum\limits_{n=0}^p c_n z^n+\sum\limits_{n=p+1}^\infty c_n z^n=\sum\limits_{n=0}^p c_n z^n $$ This means that $f$ is a polynimial with $\deg(f)\leq p$.
For your particular problem $\varphi(R)=\log(1+R)$, and it is easy to check that $$ \lim\limits_{R\to+\infty}\frac{\varphi(R)}{R}=0 $$ Hence, $f(z)=c_0$ is a constant function. Moreover, $$ |c_0|=|f(0)|\leq\log(1+|0|)=0 $$ so $c_0=0$ and $f(z)=0$ for all $z\in\mathbb{C}$.