**EDIT 2015.07.15** I believe there is no such integer $x$. See below for the rest of the edit. **END EDIT 2015.07.15**

I had some initial thoughts which seemed promising. They do not lead to a proof, but (with notation below) may show that $m$ must be very large indeed for even $m_1$ and $m_2$ to be multiples of $m$. I suspect $m_3$ will never be a multiple of $m$ in such an unlikely situation.

I prefer to use $m$ for $x$: this is because $\sigma(m) = 0 \bmod m$ means $m$ is a multiply perfect number. To avoid trivialities I assume $m \gt 1$.

Note that $\sigma(m)/m$ is bounded above by the product over all primes $q$ dividing $m$ of $q/(q-1)$, which is $O(\log p)$ with $p$ the largest prime divisor of $m$, and strictly less than $p$ when $p \gt 3$. (Indeed, it is less than $\omega(m)$, the number of distinct prime divisors of $m$, when $\omega(m) \gt 4$.) While $\sigma(m)$ itself does not have to be multiperfect, I suspect there are finitely many numbers with $\sigma(\sigma(m))$ a multiple of $m$. In particular, let $g_0=m_0=m$, $m_{n+1}=\sigma(m_n)$, and $g_{n+1}=\gcd(g_n,m_{n+1})$. I suspect $g_3 \lt m$. I base this suspicion on the observation that the power of $2$ exactly dividing $m_n$ appears to change between $m_n, m_{n+1}$, and $m_{n+2}$.

Let me write $w$ for $\omega(m)$ and let us note that for a multiperfect $m$, $m_1$ will
differ from $m$ by having less than $O(\log(w))$ additional prime factors, some in common with the prime factors of $m$. So the prime factorization of $m_1$ looks very much like the prime factorization of $m$.

I would like to argue that the prime factorization of $m_2$ should be quite different from that of $m_1$, because any additional powers of $q$ for $q$ a small prime dividing $n$ will affect $\sigma(q^n)$ and thus remove some prime factors. However, it is possible that there are (insanely large) odd multiperfect numbers which would be factors of $m$ and not be affected by this. Indeed, this question may be equivalent to the question of the existence of large odd multiperfect numbers.

**EDIT (Part II) 2015.07.15**
Let us look at $S(m)= \sigma(m)/m$. Letting the following products run over the distinct
primes $q$ dividing $m$, we have $\prod (q+1)/q \leq S(m) \lt \prod q/(q-1)$. (And the lower bound is at least half the upper bound, so we have $S(m) \simeq \prod q/(q-1)$.) As observed above, $S(m) \lt \omega(m)$ when $4 < \omega(m)$ and $S(m) \lt 2\omega(m)$ the rest of the time. So $S(m)$ is pretty small compared to $m$ and often small compared to $\log p$ where $p$ is the greatest prime factor of $m$.

Let $r_n = m_{n+1}/m_n = S(m_n)$. The assumption in the problem implies $r_n \gt r_0$,
for if $m$ properly divides $k$ then $S(m) \lt S(k)$. Then $m_n \gt mr_0^n$ for all $n$, since $m_n$ is an increasing sequence. I believe we can't have both conditions hold indefinitely. However, I now switch ground on
my assertion above that $g_3 < m$ always happens: I think it can, I just don't think we will see an example with fewer than a 1000 decimal digits (which isn't insanely large).

Gerry Myerson's example from the comment is thought provoking: When $S(m)$ is coprime to $m$, we clearly have $m \mid m_2$ as well as $m \mid m_1$, and by multiplicativity of $S$ we also have $S(S(m)m)=S(S(m))S(m)$. What if $S(m)$ is not coprime to $m$?
We still have $S(S(m)m) \lt S(m)S(S(m))$.

This last is the crux. $S()$ grows slower than $\log()$, so we need to show this
contradicts the growth rate implied by all $m_n$ being multiples of $m$. It is this
that inspires my confidence above, and also has me reverse my stance on this being
equivalent to odd multiperfect numbers. I hope to finish this argument in a future
edit.
**END EDIT(Part II) 2015.07.15**

Gerhard "Doesn't Have A Good Closer" Paseman, 2015.07.14