Solution?

An appealing infinite descent it seems. I took cases for divisibility of 61, and it looks like infinite descent is related. I am not sure about it, so not posting here

All solutions in positive integers of the Diophantine equation $(1) \;\; \frac{1}{x^2} + \frac{y}{xz} + \frac{1}{z^2} = \frac{1}{2013}$ is given by $(x,y,z) = (2013d, 2013d^2-2,2013d)$, where $d$ is a positive integer. Proof: By multiplying equation (1) with $2013(xz)^2$ we obtain $(2) \;\; 2013(x^2 + xyz + z^2) = (xz)^2$. According to equation (2) there is a positive integer $a$ s.t. $(3) \;\; xz = 2013a$, $(4) \;\,x^2+xyz+z^2 = 2013a^2$. Hence ${\textstyle z = \frac{2013a}{x}}$ by equation (3), which inserted in equation (4) result in $(5) \;\; x^4 + 2013ax^2y + 2013^2a^2 = 2013a^2x^2$. Consequently $2013 | x^4$, yielding $2013 \, | \, x$ since $2013 = 3 \cdot 11 \cdot 61$ is squarefree. Hence $x=2013b$, yielding $a=bz$ by equation (3). Making the substitutions $a=bz$ and $x=2013b$ in equation (5), the result is $(6) \;\; 2013^2b^2 + 2013byz + z^2 = 2013b^2z^2$. Hence $2013 \, | \, z^2$ by equation (6), implying $2013 \, | \, z$, i.e. $z=2013c$, which inserted in equation (6) yields $(7) \;\; b^2 + bcy + c^2 = 2013b^2c^2$. Let $d=GCD(b,c)$. Hence there exist two coprime positive integers $s$ and $t$ s.t. $(b,c) = (ds,dt)$, which inserted in equation (7) give us $(8) \;\; s^2 + sty + t^2 = 2013d^2s^2t^2$. Therefore $s \, | \, t^2$ and $t \, | \, s^2$ by equation (8), implying $s=t=1$ since $GCD(s,t)=1$. Consequently $y = 2013d^2-2$ by equation (8) and $x=2013b=2013c=z=2013d$ since $b=c=d$. Summa summarum, the positive integer solutions of equations are $(x,y,z) = (2013d,2013d^2-2,2013d)$, where $d$ is a positive Integer. q.e.d.