The short answer
$\newcommand{\modN}[1]{#1\,\operatorname{mod}\,N}\newcommand{\on}[1]{\operatorname{#1}}$Quantum Computers are able to run subroutines of an algorithm for factoring, exponentially faster than any known classical counterpart. This doesn't mean classical computers CAN'T do it fast too, we just don't know as of today a way for classical algorithms to run as efficient as quantum algorithms
The long answer
Quantum Computers are good at Discrete Fourier Transforms.
There's a lot at play here that isn't captured by just "it's parallel" or "it's quick", so let's get into the blood of the beast.
The factoring problem is the following: Given a number $N = pq$ where $p,q$ are primes, how do you recover $p$ and $q$? One approach is to note the following:
If I look at a number $\modN{x}$, then either $x$ shares a common factor with $N$, or it doesn't.
If $x$ shares a common factor, and isn't a multiple of $N$ itself, then we can easily ask for what the common factors of $x$ and $N$ are (through the Euclidean algorithm for greatest common factors).
Now a not so obvious fact: the set of all $x$ that don't share a common factor with $N$ forms a multiplicative group $\on{mod} N$. What does that mean? You can look at the definition of a group in Wikipedia here. Let the group operation be multiplication to fill in the details, but all we really care about here is the following consequence of that theory which is: the sequence
$$ \modN{x^0}, \quad\modN{x^1}, \quad\modN{x^2}, ... $$
is periodic, when $x,N$ don't share common factors (try $x = 2$, $N = 5$) to see it first hand as:
$$\newcommand{\mod}[1]{#1\,\operatorname{mod}\,5}
\mod1 = 1,\quad
\mod4 = 4,\quad
\mod8 = 3,\quad
\mod{16} = 1.
$$
Now how many natural numbers $x$ less than $N$ don't share any common factors with $N$? That is answered by Euler's totient function, it's $(p-1)(q-1)$.
Lastly, tapping on the subject of group theory, the length of the repeating chains
$$ \modN{x^0}, \quad\modN{x^1}, \quad\modN{x^2}, ... $$
divides that number $(p-1)(q-1)$. So if you know the period of sequences of powers of $x \mod N$ then you can start to put together a guess for what $(p-1)(q-1)$ is. Moreover, If you know what $(p-1)(q-1)$ is, and what $pq$ is (that's N don't forget!), then you have 2 equations with 2 unknowns, which can be solved through elementary algebra to separate $p,q$.
Where do quantum computers come in? The period finding. There's an operation called a Fourier transform, which takes a function $g$ written as a sum of periodic functions $a_1 e_1 + a_2 e_2 ... $ where $a_i$ are numbers, $e_i$ are periodic functions with period $p_i$ and maps it to a new function $\hat{f}$ such that $ \hat{f}(p_i) = a_i$.
Computing the Fourier transform is usually introduced as an integral, but when you want to just apply it to an array of data (the Ith element of the array is $f(I)$) you can use this tool called a Discrete Fourier Transform which amounts to multiplying your "array" as if it were a vector, by a very big unitary matrix.
Emphasis on the word unitary: it's a really arbitrary property described here. But the key takeaway is the following:
In the world of physics, all operators obey the same general mathematical principle: unitarity.
So that means it's not unreasonable to replicate that DFT matrix operation as a quantum operator.
Now here is where it gets deep an $n$ Qubit Array can represent $2^n$ possible array elements (consult anywhere online for an explanation of that or drop a comment).
And similarly an $n$ Qubit quantum operator can act on that entire $2^n$ quantum space, and produce an answer that we can interpret.
See this Wikipedia article for more detail.
If we can do this Fourier transform on an exponentially large data set, using only $n$ Qubits, then we can find the period very quickly.
If we can find the period very quickly we can rapidly assemble an estimate for $(p-1)(q-1)$
If we can do that fast then given our knowledge of $N=pq$ we can take a stab at checking $p,q$.
That's whats going on here, at a very high level.
