Let $d = GCD(y,z)$. Then there are two coprime positive integers $r$ and $s$ s.t. $y=rd$ and $z=sd$, which inserted in the original equation $(1) \;\; z(xz + 1)^2 = (5z + 2y)(2z + y)$ result in $(2) \;\; s(dsx + 1)^2 = d(5s + 2r)(2s+ r)$. Hence $d|s$ by (2), i.e. $s = td$ for a positive integer $t$, which inserted in (2) give us $(3) \;\; t(d^2tx + 1)^2 = (5dt + 2r)(2dt + r)$. According to (3) we have $t | 2r^2$, yielding $t | 2$ since $1 = GCD(r,s) = GCD(r,td)$ implies $GCD(r,t)=1$. Next assume $t=2$. Inserting this value of $t$ in (3), we obtain $(4) \;\; (2d^2x + 1)^2 = (5d + r)(4d + r)$. If $p$ is a prime divisor of $5d + r$ and $4d + r$, then $p \mid (5d + r) - (4d + r) = d$ and $p \mid 5(4d + r) - 4(5d + r) = r$, implying $p \mid GCD(r,d)=1$. This contradiction means $GCD(5d+r,4d+r) = 1$. Hence by (4) $(5) \;\; 5d + r = u^2$, $(6) \;\; 4d + r = v^2$, $(7) \;\; 2d^2x + 1 = uv$, where $u$ and $v$ are coprime positive integers. By (5)-(6) we obtain $u^2 - v^2 = (5d + r) - (4r + r) = d$, which combined with (7) give us $uv > 2d^2x > d^2 = (u^2 - v^2)^2 = (u - v)^2 \cdot (u + v)^2 \geq 1^2 \cdot 4uv = 4uv$. Consequently $t \neq 2$ by contradiction. Hence $t=1$ (since $t|2$), which inserted in (3) result in $(8) \;\; (d^2x + 1)^2 = (5d + 2r)(2d + r)$. Moreover $z = sd = td^2 = d^2$. If $d$ is even, then the LHS of (8) is odd and the RHS of (8) is even (since $5d + 2r$ is even). Therefore $d$ is not even, i.e. $z=d^2$ is an odd perfect square. q.e.d.