I have to prove that if $u:\mathbb{R}^n\to\mathbb{R}$ is harmonic and: \begin{equation} \dfrac{|u(x)|}{\sqrt{1+|x|}}\leq C \ \ \ \forall x\in\mathbb{R^n} \end{equation} then $u$ is constant.
MY ATTEMPT: cosider $R>0$ and $B_R(0)$ the ball of radius $R$ centered at $0$. Then $\forall x\in B_R(0)$: \begin{equation} |u(x)|\leq C\sqrt{1+|x|}\leq C\sqrt{1+R}=C' \end{equation} Hence on $B_R(0)$ we have that $-C'\leq u(x)\leq C'$ which means that $u$ is bounded from below. Using now Liouville's theorem we can conclude that u is constant on $B_R(0)$ and finally for the arbitrariety of $R$ we can conclude that $u$ is constant on the whole $\mathbb{R}^n$
EDIT: Using the face that $u$ is harmonic we have from the mean value property that for $x$ fixed and for any $r>0$: \begin{equation} u(x)=\dfrac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)} u(y)dS(y) \end{equation}
from which dividing for $\sqrt(1+r)$ on both sides we have: \begin{equation} \dfrac{u(x)}{\sqrt{1+r}}=\dfrac{1}{|\partial B_r(x)|\sqrt{1+r}}\int_{\partial B_r(x)} u(y)dS(y)=\dfrac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)} \dfrac{u(y)}{\sqrt{1+|y|}}dS(y) \end{equation} Now using repeating this argument we get: \begin{equation} \dfrac{|u(x)|}{\sqrt{1+r}}\leq\dfrac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)} \dfrac{u(y)}{\sqrt{1+|y|}}dS(y)\leq C \end{equation} where in the last I have used the Hypothesis of the exercise. At this point we have find that: \begin{equation} -C\sqrt{1+r}\leq u(x)\leq C\sqrt{1+r} \end{equation} which gives a uniform bound for $u(x)$ because of the arbitrariety of $x$.Thus applying Liouville theorem $u$ is harmonic and bounded from below and hence constant.
Here is a way that builds on your idea of showing that $u$ is bounded. The principle of spherical means tells us that for harmonic $u$, $$u(x) = \frac{1}{4\pi r^2}\int_{\partial B(x,r)}u(y)dS(y).$$ That is, $u(x)$ is equal to its average value over the sphere (of any radius) about $x$. Now, on the sphere, $|x| = r$ so that $$\frac{u(x)}{\sqrt{1+r}} = \frac{1}{4\pi r^2}\int_{\partial B(x,r)}\frac{u(y)}{\sqrt{1+|y|}}dS(y).$$ Therefore, $$-C\sqrt{1+r} \leq u(x) \leq C\sqrt{1+r} \quad \forall r>0.$$ For any fixed $r$, we achieve global boundedness because the point $x$ is arbitrary.