Solution:
Call $a=1+\frac{x}{y+z}$, $b=1+\frac{y}{z+x}$, $c=1+\frac{z}{x+y}$ to get
$$a^{2}+b^{2}+c^{2}=\frac{27}{4}$$
Since it is also true that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$$
the quadratic-harmonic means inequality implies
$$\frac{3}{2}=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{3}{2}$$
So the inequality in the middle holds as an equality, and this happens whenever $a=b=c$, from which $1+\frac{x}{y+z}=1+\frac{y}{z+x}=1+\frac{z}{x+y}$.
But $1+\frac{x}{y+z}=1+\frac{y}{z+x} \Leftrightarrow x^{2}+x z=y^{2}+y z \Leftrightarrow(x-y)(x+y)=z(y-x)$ and the two sides of this equality will be of different sign, unless $x=y$ in which case both sides become $0$. So $x=y$, and similarly $y=z$, thus $x=y=z$.
Indeed, any triad of equal natural numbers $x=y=z$ is a solution for the given equation, and so these are all its solutions.

Solution 2:
The given equation is equivalent to
$$\frac{27}{4}=(x+y+z)^{2}\left(\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}+\frac{1}{(x+y)^{2}}\right)$$
Now observe that by the well known inequality $a^{2}+b^{2}+c^{2} \geq ab+bc+ca$, with $\frac{1}{y+z}$, $\frac{1}{z+x}$, $\frac{1}{x+y}$ in place of $a, b, c$; we get:
$$\begin{aligned} \frac{27}{4} & =(x+y+z)^{2}\left(\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}+\frac{1}{(x+y)^{2}}\right) \\ & \geq (x+y+z)^{2}\left(\frac{1}{(y+z)(z+x)}+\frac{1}{(z+x)(x+y)}+\frac{1}{(x+y)(y+z)}\right) = \frac{2(x+y+z)^{3}}{(x+y)(y+z)(z+x)} \\ & = \frac{(2(x+y+z))^{3}}{4(x+y)(y+z)(z+x)} = \frac{((x+y)+(y+z)+(z+x))^{3}}{4(x+y)(y+z)(z+x)} \stackrel{\text{AM-GM}}{\geq} \frac{(3 \sqrt[3]{(x+y)(y+z)(z+x)})^{3}}{4(x+y)(y+z)(z+x)} . \\ & = \frac{27}{4} \end{aligned}$$
This means all inequalities in the above calculations are equalities, and this holds exactly whenever $x+y=y+z=z+x$, that is $x=y=z$. By the statement's demand we need to have $a, b, c$ integers. And conversely, any triad of equal natural numbers $x=y=z$ is indeed a solution for the given equation, and so these are all its solutions.