Well the standard techniques would take advantage of the fact that the equation doesn't explicitly involve the independent variable $x$ to integrate the equation once, thereby leading to the conservation law
$$
y^2 + 2(1+y'^2)^{-1/2} = r^2,\tag1
$$
where, $r>0$ is a constant. Note that we must have $|y| < r$

The relation (1) can then be solved to yield, after separation of variables,
$$
\frac{(r^2-y^2)\,dy}{\bigl((2-r^2+y^2)(2+r^2-y^2)\bigr)^{1/2}} = dx.\tag2
$$
The left hand integral can be computed in terms of elliptic functions, of course, but you don't need to do this to do a qualitative analysis.

If $r>\sqrt2$, then we must have $|y|\ge\sqrt{r^2-2}$, so, replacing $y$ by $-y$, we can assume that $\sqrt{r^2-2}\le y(x) < r$. Because the integral on the left hand side of (2) over the interval $\sqrt{r^2-2}\le y \le r$ is finite, it follows that $y(x)$ can only be defined over an $x$-interval of finite length. Thus, there is no entire solution in this case.

If $r = \sqrt2$, then, again, we could only have $y=0$ where $y'=0$. However, the left-hand integral of (2) over the interval $0<y<a$ when $0<a<\sqrt{2}$ is infinite, so we cannot have $y$ vanishing for any finite value of $x$, thus, $y$ cannot vanish and we can assume $y>0$. The above integral then leads to a solution $y(x)$ on an interval of the form $(-\infty,C]$ where $C<\infty$, so there is no entire solution in this case either.

Finally, if $r<\sqrt{2}$, then we have $|y| < r$, but the integral of the lefthand side of (2) over the interval $-r < y < r$ is finite, so, again, there is no solution defined over the entire $x$-axis.