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Is it possible to solve for x this kind of equation? Since 1,2,3 are not multiples to each other I see a priori no possibility.
$$1^x+2^x+3^x=0; x?$$

Computing this on Wolfram Alpha, for example, returns two complex solutions for $x$. Wolfram Alpha usually returns a formal and algebraical expression to represent the solution, but this time no formal one is returned, instead, just a long, infinite and complex number with seems to be the result of an iterating computation rather than a formal way to solve it.

It certainly can't have any real value solutions since,
$$1^x+2^x+3^x=0$$
$$1+2^x+3^x=0$$
$$2^x+3^x=-1$$
And $a^x>0$ for all real $x$ and all real $a>0$.

So it makes sense that if you want a solution, the search needs the extended number system of complex numbers.

If you ask Wolfram Alpha to solve
$$2^x+3^x=-1$$
it will give you the numerical solutions
$$x=-0.454397 pm 3.59817i$$

If you want to try and obtain these by hand you could try applying the numerical Newton Raphson method to the equation; it works with complex numbers as well as the reals, but the arithmetic may be daunting especially if you are interested in solutions to
$$sum_k=1^n k^x=0$$

To explore this properly would need software that can apply the Newton Raphson method in complex numbers and show you what it's doing.

A quick google search makes me think Maxima may do it for free : http://maxima.sourceforge.net/
or GNU Octave : http://www.gnu.org/software/octave/
although I have no experience of either, or even if they can cope with complex numbers.

The irony is, of course, that you ideally want to start near a solution. I see this is becoming an interesting project; be good to graph the modulus of the complex function for small values of k, as often done for Zeta function.

Let us know if you manage it !

Too long for a comment.

After @Martin Hansen's answer, I had a look to he zero of equation
$$f_n(x)=sum_k=1^n k^x$$
(give a lot of attention to @reuns's comment).

I used Newton method. For the base case $(n=3)$ starting with $x_0=-1+pi,i$ the iterates are

$$left(
beginarraycc
1 & -0.501820 ,+ ,3.86512 ,i \
2 & -0.487049 ,+ ,3.58176,i \
3 & -0.454116 ,+ ,3.59875 ,i \
4 & -0.454397 ,+ ,3.59817, i
endarray
right)$$

Below are given some results which could be a good start (I hope) for a deeper exploration (as @Martin Hansen suggested).
$$left(
beginarraycc
3 & -0.454397, + ,3.59817 , i \
4 & -0.625971, + ,3.12712 , i \
5 & -0.714285, + ,2.83349 , i \
6 & -0.767633, + ,2.62901 , i \
7 & -0.803209, + ,2.47644 , i \
8 & -0.828584, + ,2.35711 , i \
9 & -0.847585, + ,2.26049 , i \
10 & -0.862348, + ,2.18022 , i \
11 & -0.874153, + ,2.11212 , i \
12 & -0.883812, + ,2.05341 , i \
13 & -0.891868, + ,2.00209 , i \
14 & -0.898692, + ,1.95672 , i \
15 & -0.904551, + ,1.91623 , i \
16 & -0.909640, + ,1.87978 , i \
17 & -0.914103, + ,1.84675 , i \
18 & -0.918051, + ,1.81661 , i \
19 & -0.921571, + ,1.78897 , i \
20 & -0.924730, + ,1.76349 , i
endarray
right)$$