Exponents are not so straightforward!

Consider the equation $\displaystyle x^{c^c}=c^x$ where $c$ is a real constant and $x$ is what we are looking forward to,

$\displaystyle \implies c^c\ln x =x\ln c \implies -e^{-\ln x}\ln x = -c^{-c}\ln c$ and Now taking Lambert W function on both sides we have

$\displaystyle -\ln x = {\rm W}(-c^{-c}\ln c) \implies x=e^{-{\rm W}(-c^{-c}\ln c)}$

For $c=1.2$ it turns out to be $x\approx20.67660773$ & $x\approx 1.190540017$ and so $x_0=2$

Now for the equation $\displaystyle y^{k}=d^y \implies k\ln y=y\ln d \implies -e^{-\ln y}\ln y=\frac{-\ln d}{k}\implies {\rm W}(-e^{-\ln y}\ln y) = {\rm W}(\frac{\ln d}{k})\implies y = e^{-{\rm W}(\frac{\ln d}{k})}$

For $k=4$ & $d=2$ since $x_0^{x_0}=4$ we have two solutions $x\approx 16$ & $x\approx 1.24$ and so $y_0=2$

Now the last equation is $\displaystyle z^2=2^z$ which handled in similar manner shows $z=2,4$ are the only positive solutions. So answer is $\boxed{2+4=6}$