How to solve exponential equations like $2^x+x=5$?

You are stuck for good reason. That equation does not have an algebraic solution. SInce $2^x + x$ is an increasing function of $x$ there will be a unique value of $x$ that makes it $5$. You can calculate it numerically as accurately as you like.

Since $\frac{\mathrm{d}}{\mathrm{d}x}\left(2^x+x\right)=\log(2)\,2^x+1\gt0$, or if it is assumed that $2^x$ is monotonically increasing, we have that $2^x+x$ is monotonically increasing for all $x\in\mathbb{R}$. Since $2^1+1=3$ and $2^2+2=6$, we see that the $x$ which satisfies $2^x+x=5$ is between $1$ and $2$. However, there is no solution in terms of elementary functions.

However, if you are willing to use special functions like Lambert W: $$\newcommand{\W}{\operatorname{W}} 2^x+x=5\\ e^{x\log(2)}=5-x\\ 32\log(2)\,e^{-(5-x)\log(2)}=(5-x)\log(2)\\ 32\log(2)=(5-x)\log(2)e^{(5-x)\log(2)}\\ \W(32\log(2))=(5-x)\log(2)\\ x=5-\frac{\W(32\log(2))}{\log(2)} $$ Approximately, $x\doteq1.7156207332755861694$.

Without using Lambert function (which I am in love with !), you can approximate the function building around $x=2$ (which would be the solution of $2^x+x=6$) the simplest $[1,1]$ Padé approximant of $$f(x)=2^x+x-5$$ This would give $$2^x+x-5 \approx \frac{1+\frac{ 1+8 \log (2)+14 \log ^2(2)}{1+4 \log (2)}(x-2)}{1-\frac{2 \log ^2(2)}{1+4 \log (2)}(x-2)}$$ Then $$x\approx \frac{1+28 \log ^2(2)+12 \log (2)}{1+14 \log ^2(2)+8 \log (2)}\approx 1.71574$$ Better would be the result using instead the $[1,n]$ approximant and get an approximation still at the price of a linear equation. For example, the next one would give $$x\approx \frac{3+292 \log ^3(2)+222 \log ^2(2)+48 \log (2)}{3+146 \log ^3(2)+132 \log ^2(2)+36 \log (2)}\approx 1.71560$$