If you substitute $v = v + u$, $h = h + u$ and $s = s + u$, the equation becomes much simpler and the variable $u$ goes away entirely! Then substituting $h = hv$ reduces the degree of $v$ down to $4$ (removing a factor of $v^4$) and following it up by $v = st$ gets rid of $s$ after removing a factor of $s^4$. So you end up with the curve $$ (16t^4 - 96t^3 + 240t^2 - 288t + 144)h^4 + (-36t^4 + 200t^3 - 480t^2 + 576t - 288)h^3 + (35t^4 - 180t^3 + 408t^2 - 480t + 240)h^2 + (-18t^4 + 86t^3 - 180t^2 + 200t - 96)h + (4t^4 - 18t^3 + 35t^2 - 36t + 16) = 0 $$ which has genus $1$. Perhaps you can do something from here.
Continued: following Mike Zieve's suggestion, let's show there are no real points. First, substituting in succession $t = t + 1$, $h= (h+1)/2$ and $t = (1+g)/(1-g)$ converts it into the nice symmetric (!) expression $$ (9g^4+6g^2+1)h^4-8gh^3+(6g^4+12g^2+2)h^2+(-8g^3-8g)h+g^4+2g^2+1 $$ which we want to show is positive everywhere. In other words, we need to show $$ (3g^2 + 1)^2h^4 + (6g^4+12g^2+2)h^2 + (g^2 + 1)^2 \geq 8gh^3 + 8(g^2+1)gh. $$ We have $$ (6g^4+12g^2+2) = 4/3 + 2/3 + 6g^4 + 12g^2 \geq 4/3 + 4g^2+ 12g^2 $$ by AM-GM. Split up the LHS as $(3g^2 + 1)^2h^4 + 4/3h^2$ plus $16g^2h^2 + (g^2 + 1)^2$. The first expression is $$ \geq 2(3g^2 + 1)|h|^3 \sqrt{4/3} >= 2\cdot2 \cdot \sqrt{3} |g| |h|^3 \cdot 2/\sqrt{3} \geq 8gh^3 $$ and the second is $$ \geq 8gh(g^2 + 1), $$ finishing the proof. I guess you still have to check that equality is never attained for rationals, and also the boundary conditions (stuff we divided by), but that is pretty straightforward.