$e$ is the only number in the universe having this property?

yes, the function $f(x)=\frac{\log(x)}{x}$, $x>0$ has a unique maximum (just look at the derivative) for $x=e$ which by taking exponential (strictly monotone) is equivalent to your statement.

$$ e\log x \leq x \log e \Leftrightarrow x^e \leq e^x $$

with strict inequality iff $x\neq e$.

Claim : Whenever , we have real numbers $a,b$ with $e\le a<b$, we have $b^a<a^b$

Proof : for real numbers $a,b> 1$ , the condition $$b^a<a^b$$ is equivalent to $$a\ln(b)<b\ln(a)$$ and therefore equivalent to $$\frac{a}{\ln(a)}<\frac{b}{\ln(b)}$$ But $f(x)=\frac{x}{\ln(x)}$ has derivate $f'(x)=\frac{\ln(x)-1}{(\ln(x))^2}$, which is positive for $x>e$ and $0$ for $x=e$. Hence, $f(x)$ is stricly increasing for $x\ge e$. This implies the claim.

This useful generalization can , in particular be used to decide whether $e^{\pi}$ or $\pi^e$ is larger.

As already pointed out in the answer above, the function $f(x)$ has even a global minimum for $b=e$, so if we set $a=e$, we have $b^e<e^b$ for $b\ne e$ and $b>0$. A number $a\ne e$ cannot have the desired property because if we choose $b=e$, we get $a^b<b^a$ contradicting the required inequality $a^b>b^a$