Why is the polynomial ring $\Bbb R[x]$ a PID but $\Bbb Z[x]$ is not? Why is the polynomial ring $\Bbb R[x]$ a PID but $\Bbb Z[x]$ is not? 
The question is asking me to prove that $\Bbb R[x]$ is a PID. I'm assuming you would go about this knowing that every field is a PID, but I'm stuck on the differences between $\Bbb R[x]$ and $\Bbb Z[x]$.
There are many ways to answer this question, at different levels, but I would say that I think the right reason is because $\mathbb{Z}$ is not a field.
I will give you 3 supporting evidences for that.

The proof that $\mathbb{R}[X]$ is a PID relies on long division of polynomials, and the fact that you can divide a polynomial by any non zero polynomial, and get a quotient and a remainder in $\mathbb{R}[X]$. This relies on the fact that any nonzero real number is invertible.

This does not work in $\mathbb{Z}[X]$: for example, if you divide $X$ by $2$, you will get a quotient which does not lie in $\mathbb{Z}[X]$. What goes wrong here is that the invertible elements of $\mathbb{Z}$ are $\pm 1$.
OK, this may be not satisfactory, because what this really proves is that $\mathbb{Z}[X]$ is not Euclidean for the degree function. It does not prove that $\mathbb{Z}[X]$ is not Euclidean for another function. Worse, there exists PID's which are not Euclidean.

Another reason (see Bernard's answer): in a PID, any non zero prime ideal is maximal.
This is not the case for $\mathbb{Z}[X]$, since $p\mathbb{Z}[X]$ is a prime ideal but not maximal, for any prime number $p$.

Note that this cannot happen in $\mathbb{R}[X]$ ( $p\mathbb{R}[X]=\mathbb{R}[X]$, since $p$ is invertible in $\mathbb{R}$).
Once again, the main difference is that you have a lot of non invertible elements in $\mathbb{Z}$. Ok, but this is still too vague. SO let's go for :

Thm. Let $A$ be a commutative ring with $1$. Then the following properties are equivalent:

i) $A$ is a field
ii) $A[X]$ is an Euclidean domain
iii) $A[X]$ is a PID.
Only iii)$\Rightarrow$ i) needs a proof: in a PID, any nonzero prime ideal is maximal.
Since $A[X]$ is a PID, it is an integral domain, and thus so is $A$ (this is a subring of $A[X]$)
Since evaluation at $0$ induces  a ring isomorphism $A[X]/(X)\simeq A$, $(X)$ is a nonzero prime ideal. Since $A[X]$ is a PID, $(X)$ is thus a maximal ideal. Consequently, $A[X]/(X)$ is a field, meaning that $A$ is a field.
Find an ideal that is not principal in $\mathbb Z[x]$. Hint: $\mathbb Z$ is a PID so your idea will need something with $x$ in it.

Show that every ideal in $\mathbb R[x]$ is principal. Hint: every number is related by multiplication.