According with the graph of the functions $f(x)=x^{x-\sqrt x}$ and $g(x)=\sqrt x+1$ the equation $$x^{x-\sqrt x}=\sqrt x+1;\space x\gt 0$$ has the two approximated solutions $x_1\approx0.216$ and $x_2\approx2.618$.
Well, with a simple trick I have got the exact value (closed form) of $x_2$ but I can't do so for $x_1$. (It clearly could be obtained several ways for $x_1$ a value in approximated form like the given $x_1\approx 0.216$).
My question is the following: Is there any closed (exact) form for $x_1$?
Here's a partial answer: if $x_1$ is algebraic, then it must be a quadratic irrational of the specific form $\frac14 + \alpha - \sqrt{\alpha}$ for some non-square rational $\alpha>0$.
Proof: Assume $x_1$ is algebraic. Then so is the RHS $\sqrt{x_1}+1$. By Gelfond-Schneider, the LHS is transcendental unless $x_1 \in \{0,1\}$ (which we know isn't the case), or else $x_1 - \sqrt{x_1}$ is rational. Therefore the latter must be true, which means $x_1$ has degree at most $2$. We now eliminate the possibility that $x_1$ is rational.
Suppose that $x_1$ is rational, then since $x_1 - \sqrt{x_1}$ is rational, so too is $\sqrt{x_1}$. Let $\sqrt{x_1} = \frac{a}{b}$ with coprime $a,b >0$. Since we know $x_1$ approximately we can verify by computation that $a,b>1$, and also that the exponent $x_1 - \sqrt{x_1}$ is negative. We thus have
$$\big(\frac{a}{b}\big)^{-Q} = \frac{a}{b} + 1 \implies \big(\frac{a}{b}\big)^{Q} = \frac{b}{a+b},$$
where $Q$ is a positive rational. This is easily seen to be impossible by the rational root theorem or by unique factorization (for instance, consider a prime $p$ dividing $b$, it must appear only on the denominator on the left but only on the numerator on the right).
It follows that $x_1$ must be a quadratic irrational. It is not hard to check that any number in the interval $[0,\tfrac14]$ for which $x_1 - \sqrt{x_1} \in \mathbb Q$ takes the form $\frac14 + \alpha - \sqrt{\alpha}$.
Perhaps someone can extend the argument much much further and eliminate the quadratic irrational case? One can alternatively use computation to prove that $x_1$ cannot be the root of a quadratic whose coefficients are bounded (in magnitude) by $C$ for various values of $C$. The ISC calculation I did above suggests that can probably take $C = 5800$, and integer relation algorithms can take this bound much higher.