\begin{align*} \text{Let's x=a-3 y=b-5 z=c-7}\\ \ x+y+z=0\\ \text{Since }\ x^3+y^3+z^2-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0 \text{, then}\\ \ xyz=180\\ \text{Afterward just check the cases.}\\ \end{align*}

What you write @above is really simple, but how we check cases.... I mean, they are thousants of them since x, y, z can also be negative!!!

Continue from $x+y+z=0, xyz=180\implies xy(x+y)=-180$. Note that $(x-y)^2=(x+y)^2-4xy=(x+y)^2+\frac{720}{x+y}$ is a perfect square and $r=x+y$ is an integer dividing $180$. If $r>0$, then we have \[r^2+\frac{720}r\geq (r+1)^2\implies 720\geq r(2r+1)\implies 5761\geq (4r+1)^2\implies r\leq 18\]However, since equality never occurs, we actually have the stronger inequality: \[r^2+\frac{720}{r}\geq (r+2)^2\implies 180\geq r(r+1)\implies r\leq 12\]$r=12$ gives $144+60=204$, $r=10$ gives $172$, $r=9$ gives $161$, $r=8$ does not divide $180$, $r=6$ gives $36+120=156$, $r=4$ gives $16+180=196$ (good), $r=3$ gives $249$, $r=2$ gives $364$, and $r=1$ gives $721$. The only good case is $x+y=4\implies xy=-45\implies (x,y)=(9,-5), (-5,9)$. If $r=x+y<0$, then we must have \[r^2+\frac{720}{r}\leq (r+1)^2\implies r(2r+1)\leq 720\implies r\geq \frac{-\sqrt{5761}-1}{4}\implies r\geq -18\]Since equality does not occur, we must have \[r^2+\frac{720}{r}\leq (r+2)^2\implies 180\geq r(r+1)\implies r\geq -13\]$r=-12$ gives $144-60=84$, $r=-10$ gives $28$, and $r=-9$ gives $1$ (good). Any lower and the expression becomes negative, and not a perfect square. If $r=x+y=-9\implies xy=20\implies (x,y)=(-4,-5), (-5,-4)$. In conclusion, the only solutions are $\boxed{(x,y)=(9,-5),(-5,9),(-4,-5),(-5,-4)}$.