It seems easy enough...

First, notice that $(x+y+z)^2=1$

$x^2+y^2+z^2+2xy+2yz+2zx=1$

$(x^2+y^2+z^2)+2(xy+yz+zx)=1$

$2+2(xy+yz+zx)=1$

$\boxed{xy+yz+zx=-\frac{1}{2}}$

Next, notice that $(x+y+z)(x^2+y^2+z^2)=2$

$x^3+y^3+z^3+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=2$

$(x^3+y^3+z^3)+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=2$

$3+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=2$

$\boxed{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=-1}$

Third, $(x+y+z)^3=1$

$x^3+y^3+z^3+3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz=1$

$(x^3+y^3+z^3)+3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)+6xyz=1$

$3+3(-1)+6xyz=1$

$\boxed{xyz=\frac{1}{6}}$

And finally, piece de resistance, $(x^3+y^3+z^3)(x+y+z)-(x^2+y^2+z^2)(xy+yz+zx)+(x+y+z)(xyz)=3(1)-2(-\frac{1}{2})+(1)(\frac{1}{6})=3+1+\frac{1}{6}=\frac{25}{6}$

$(x^4+y^4+z^4+x^3y+x^3z+y^3x+y^3z+z^3x+z^3y)-(x^3y+x^3z+y^3x+y^3z+z^3x+z^3y+x^2yz+xy^2z+xyz^2)+(x^2yz+xy^2z+xyz^2)=\frac{25}{6}$

$(x^4+y^4+z^4)+(x^3y+x^3z+y^3x+y^3z+z^3x+z^3y-x^3y-x^3z-y^3x-y^3z-z^3x-z^3y)+(-x^2yz-xy^2z-xyz^2+x^2yz+xy^2z+xyz^2)=\frac{25}{6}$

$(x^4+y^4+z^4)+(0)+(0)=\frac{25}{6}$

$\boxed{x^4+y^4+z^4=\frac{25}{6}}$

Let $P_n = x^n+y^n+z^n$ , where $n$ is a positive integer, and $S_1 = x+y+x = P_1$ , $S_2 = xy+yz+zx$ , and $S_3 = xyz$ . By Newton's identities or Newton's sums , we have:

$\begin{aligned} P_1 & = S_1 = 1 & \small \blue{\text{Given}} \\ P_2 & = S_1P_1 - 2S_2 = 1-2S_2 = 2 & \small \blue{\implies S_2 = -\frac 12} \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 2+\frac 12 + 3S_3 = 3 & \small \blue{\implies S_3 = \frac 16} \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 3+1 + \frac 16 = \frac {25}6 \end{aligned}$

Therefore, $m+n = 25+6 = \boxed{31}$ .

From the given equations we get $x+y+z=1, xy+yz+zx=-\dfrac{1}{2}, xyz=\dfrac{1}{6}$ . So $x, y, z$ are the roots of the equation $X^3-X^2-\dfrac{1}{2}X-\dfrac{1}{6}=0$ , or $X^4=X^3+\dfrac{1}{2}X^2+\dfrac{1}{6}X$ . Therefore $x^4+y^4+z^4=3+1+\dfrac{1}{6}=\dfrac{25}{6}$ . So the required answer is $25+6=\boxed {31}$