Solve $3^x + 28^x=8^x+27^x$ The equation $3^x + 28^x=8^x+27^x$ has only the solutions $x=2$ and $x=0$? If yes, how to prove that these are the only ones? 

Since for all sufficiently large $x$ the expression $28^x + 3^x - 8^x - 27^x$ is positive and increasing, there cannot be roots $x$ beyond that point.
The question is how to pin down where the expression becomes positive and increasing.
Now $28^x$ grows fastest, by a factor of $28$ when $x$ is incremented by 1.  This suggests a strategy of dividing through by $28^x$ to get a simpler expression to estimate.
Define $f(x) = 1 + \left(\frac{3}{28}\right)^x - \left(\frac{8}{28}\right)^x - \left(\frac{27}{28}\right)^x$.  As noted in the Question, $f(2) = 0$.  By standard calculus techniques:
$$ f'(x) = -\left( \log \frac{28}{3} \right)\left(\frac{3}{28}\right)^x
 + \left( \log \frac{7}{2} \right)\left(\frac{2}{7}\right)^x
 + \left( \log \frac{28}{27} \right)\left(\frac{27}{28}\right)^x $$
If $x \ge 2$, we can prove the derivative is positive:
$$ f'(x) = \left(\frac{2}{7}\right)^x \left[ \left(\log \frac{7}{2}\right)
 - \left(\log \frac{28}{3}\right)\left(\frac{3}{8}\right)^x \right]
 + \left(\log \frac{28}{27}\right)\left(\frac{27}{28}\right)^x $$
$$ \ge \left(\frac{2}{7}\right)^x \left[ \left(\log \frac{7}{2}\right)
 - \left(\log \frac{28}{3}\right)\left(\frac{3}{8}\right)^2 \right]
 + \left(\log \frac{28}{27}\right)\left(\frac{27}{28}\right)^x $$
and by direct computation the quantity in square brackets is positive, approx. $0.938664$.
Thus $f(x) \gt 0 $ for all $x \gt 2$.  After checking $f(1)$ we know the only nonnegative integer solutions are the two identified in the Question, $x=0,2$.

As clarified, however, we need to determine all real solutions $x$.  We can do this almost by inspection, calling upon an extension of Descartes Rule of Signs due to Laguerre(1883).  Setting $z=3^x \gt 0$ we can rewrite the problem as finding positive real roots of the function:
$$ g(z) = z^{\log_3 28} - z^3 - z^{\log_3 8} + z $$
The number of sign changes is two, and by the version Thm. 3.1 in the PDF linked above, there are at most two positive real roots.  Since we know $z=1,9$ are roots, these are the only positive real roots (from which it follows $x=0,2$ are the only real roots of the original problem).
It should be evident from this brief discussion that a much broader application of the principle can be made, e.g. like the generalization @IvanLoh has proved.
