Are those two numbers transcendental?

$$u^u=ve^v=\pi\Rightarrow u^u=e^{v+\ln v}=\pi\Rightarrow\begin{cases}u\approx1.85411\\v\approx 1.07364\end{cases}$$ The function $f(x)=x^x$ is injective for $x\gt 1$ and so is $g(x)=xe^x$ for $x\gt 0$ hence the finded $u$ and $v$ are unique real solutions.

Besides $u$ cannot be rational because if $$x=\left(\frac ab\right)^{\frac ab}\iff b^ax^b-a^a=0$$ then $x$ is algebraic but we know that $\pi$ is trascendental. It follows that $u$ is algebraic irrational or trascendental.

On the other hand, Geldfond-Schneider theorem ensures that if $a$ and $b$ are algebraic with $a\notin\{0,1\}$ and $b$ irrational then $a^b$ is trascendental so it is not impossible that $u$ would be algebraic but it could be trascendental also. At the current state of knowledge on this topic, we can not guarantee something more on the nature of $u$ besides of it is not rational. We only know that $u$ is algebraic of degree greater than $1$ or trascendental, nothing more.

Instead, $v$ could be rational provided $e^v$ is trascendental and we ran into the same impasse. I stop here.