Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$

Consider the function for $x>1,n>1$

$$f(n)=n^x-(n-1)^x$$ $$f'(n)=xn^{x-1}-x(n-1)^{x-1}=x(n-1)^{x-1}[(1+\frac{1}{n-1})^{x-1}-1]>0$$

So $f(n)$ is increasing for $x>1,n>1$.

Now rewrite the equation $$5^x+7^x+11^x=6^x+8^x+9^x$$ $$\color{red}{(11^x-10^x)}+\color{blue}{(10^x-9^x)}=\color{red}{(8^x-7^x)}+\color{blue}{(6^x-5^x)}$$ Comparing the red and blue parts, LHS is larger than RHS due to increasing $f(n)$.

So the equation holds only if $0\le x\le 1$ for natural number solutions, i.e. $x=0$ or $x=1$.

Let $f(x)=x^k,$ where $k>1$ or $k<0$.

Thus, $f$ is a convex function and since $(11,7,5)\succ(9,8,6),$ by Karamata we obtain: $$f(11)+f(7)+f(5)>f(9)+f(8)+f(6).$$

Also, for $0<k<1$ we see that $f$ is a concave function.

Thus, by Karamata again $$f(11)+f(7)+f(5)<f(9)+f(8)+f(6).$$ Thus, it remains to check, what happens for $k\in\{0,1\}$.

Another elemenatary solution, using that $x$ is supposed to be a natural number. It even uses a technique the OP considered:

The equation is equivalent to

$$\left(\frac{5}{11}\right)^x + \left(\frac{7}{11}\right)^x +\left(\frac{11}{11}\right)^x = \left(\frac{6}{11}\right)^x +\left(\frac{8}{11}\right)^x +\left(\frac{9}{11}\right)^x$$

All terms are positive and the left hand side contains a summand 1 in the form of $\left(\frac{11}{11}\right)^x$.

From the terms on the right hand side, $\left(\frac{9}{11}\right)^x$ is the biggest, but it will of course still decrease for increasing $x$. Using a calculator will show you that $\left(\frac{9}{11}\right)^6 < \frac13$.

That means for $x \ge 6$, the right hand side consists of the sum of 3 values, the highest of which is less than $\frac13$. That means the right hand side is less than 1, while the left hand side is bigger than 1, leading to a contradiction.

Now 'only' the cases $x=2,3,4,5$ need to be checked by hand, and they don't lead to the equation being fullfilled.