Question: With additional parameters that: \begin{align} \ r^2 &= x\_j x\_j \\ \end{align} and \begin{align} \ \frac{\partial x\_i}{\partial x\_j} = \delta\_{ij}\\ \end{align}

So, what I managed to figure so far is that: \begin{align} \nabla ^2 r^n &= \nabla \cdot (\nabla r^n ) \\ 
&=\nabla \cdot (nr^{n-1} \frac{\partial r}{\partial x\_i}) \\
\end{align}

where I found that: \begin{align} \ \frac{\partial r}{\partial x\_i} = \frac{x\_i}{r} \\ \end{align}
using the above parameters, but beyond that, I am stuck on how to proceed with the rest of the equation for the proof.

Answer: 
In your solution, I believe you should have written:
$$P(r) = (r-1)Q(r) + k$$
With that adjustment, the solution works perfectly. To generalize, you could try to show:
$$(x-y) \mid (P(x)-P(y))$$
(To do this, you could start out by showing it for $P(x)=x^n$, and then generalize it to all polynomials). Once you do this, by substituting $y=0,1$), you can see that for all $x \in \mathbb{Z}$:
$$x \mid (P(x)-P(0))$$
$$(x-1) \mid (P(x)-P(1))$$
Since one of $x$ and $x-1$ is even, one of $P(x)-P(0)$ and $P(x)-P(1)$ must be even. Since both $P(0)$ and $P(1)$ are odd, $P(x)$ must be odd as well. Hence, $P(x) \neq 0$ for any $x \in \mathbb{Z}$. This approach can be generalized as you wanted, give it a shot!
Suppose $P(n)=0$ for some integer $n$. Then $x-n$ is a factor of $P(x)$, and we can write
$$P(x)=(x-n)Q(x)$$
for some polynomial $Q(x)$ with integer coefficients. Putting $x=0$ gives
$$-nQ(0)=P(0)$$
so $n$ must be odd because $P(0)$ is odd. And putting $x=1$ gives
$$(1-n)Q(1)=P(1)$$
so $n-1$ must be odd because $P(1)$ is odd. Contradiction.