is there any way to obtain the private key and consequently be able to decrypt the other ciphertexts?

Sometimes. A fuller answer involves number theory (but to understand why RSA works, you really need to know the basics anyways).

First off, a definition: the *order* of a value $a$ (modulo $p$) is the smallest positive integer $x$ such that $a^x \equiv 1 \pmod p$. The value $a \equiv 0 \pmod p$ doesn't have an order (it's not considered an element in the multiplicative group $\mathbb{Z}_p^*$); all other values have an order (assuming $p$ is prime).

Now, if we have $m^e = m \pmod {pq}$ (where $pq=n$, expressed as its prime factors), then we know that:

Either $m = 0 \pmod p$, or the order of $m$ modulo $p$ is a divisor of $e-1$

Either $m = 0 \pmod q$, or the order of $m$ modulo $q$ is a divisor of $e-1$

Now, if either $m = 0 \pmod p$ or $m = 0 \pmod q$, then a simple computation $\gcd(m, n)$ reveals the factorization (unless both are true, in which case we have $m = 0$; we can't deduce anything from that).

The other case is if both have orders; if the two orders are different, say, $x$ and $y$ (with $x < y$), then a simple computation of $\gcd( m^x - 1, n )$ reveals the factorization.

On the other, if the two orders are the same, well, the knowledge of $m$ doesn't reveal a factorization.

And, because $e$ is typically small, $e-1$ is easy to factor, hence it is practical to test $\gcd( m^x - 1, n )$ for all divisors $x$ of $e-1$

Now, there are some simplifications: to test if $m$ is order 1 (either modulo $p$ or $q$), we can just compute $\gcd( m - 1, n )$ (and if both sides have that order, that corresponds to $m = 1$)

And, for the order $2$, it turns out we can compute $\gcd( m + 1, n )$ (and if both sides have that order, that corresponds to $m = n-1$)

In most cases, computing $\gcd(m, n), \gcd(m-1, n), \gcd(m+1, n)$ is sufficient to recover the factorization; however the fuller test will catch some more cases.