A related problem. Assume the point that gives the nearest distance that lies on the curve is $(x,y)$, so the distance between the soldier and the plane is
$$ d=\sqrt{(x-3)^2+(y-7)^2}= \sqrt{(x-3)^2+x^4}$$
$$ \implies d' = \frac{ 4x^3 + 2(x-3) }{2\sqrt{(x-3)^2+x^4}}=0 \implies x=1.$$
Checking the second derivative at $x=1$ gives $d''>0$ which means it is minima. So, $y=x^2+7=8$ and the minimum distance will be at the point $(x,y)=(1,8)$ and it is equal to $d=\sqrt{5}$.