I was asked by an high school student if there is an algebraic way to find the exact value of the solution to the equation \begin{equation}\label{eq} x^{x+1}=(x+1)^x \end{equation} Let us define that with the expression "algebraic way" the student really means "the solution $x$ to the equation is an algebraic number". Now my feeling is that $x$ has to be transcendental but I'm not able to see how to prove it. Note first that there is a unique solution $2<x<3$, more precisely with Wolfram Alpha one can check that $x \approx 2.29...$ I tried to see if I can use a Gelfond-Schneider type of argument but if I write $$x=(x+1)^{\frac{x}{x+1}}$$ then $x$ will be transcendental if I know that $x+1$ is algebraic, but $x+1$ being algebraic is the same as $x$ being algebraic and then I'm stucked. I also tried some Liouville bound on a suitable approximation of $x$ with fractions but I'm not able to control the error coming from the denominators.

One can also take the logarithm of both sides of the equation yielding $$(x+1)\cdot \ln(x)=x \cdot \ln(x+1)$$ I finally tried to use Baker's theorem on linear independence over $\mathbb{Q}$ and $\overline{\mathbb{Q}}$ of logarithms but I'm also stucked because a solution would yield a linear dependence of the logarithm over $\overline{\mathbb{Q}}$ for example, making the theorem impossible to use. Maybe I think that if I'm able to find an expression $$\frac{\ln(x+1)}{\ln(x)}=\ln(f(x,y))$$ where $f(x,y)$ is a solution of a suitable polynomial $P(t) \in \mathbb{Q}[x,y][t]$ then I can use a Baker type argument, but I'm not able to find such an explicit expression for $f(x,y)$.

So I've decided to ask here if some of you have a way to prove or disprove the transcendence of $x$. Thanks in advance.