I’m assuming that $x$ is given, and you want to find $m$ so that $x^2 + (2m+1)x + m^2 - 1 = 0$. To do this, just treat the equation as a quadratic in $m$, with $x$ as a constant, and rewrite it as $m^2 + 2xm + (x^2 +x - 1) = 0$. The quadratic formula now gives $\begin{align*}m &= \frac12\left(-2x\pm\sqrt{4x^2 - 4(x^2+x-1)} \right)\\ &= -x\pm\sqrt{1-x} \end{align*}.$ Alternatively, it’s easy to complete the square: $m^2 + 2xm + (x^2 +x - 1) = (m+x)^2 + x - 1$, which gets you to $m = -x \pm \sqrt{1-x}$ even faster.
Added: It occurs to me that what’s wanted might be the range of values of $m$ for which there is a real $x$ making $y=0$. Let $f(x) = -x + \sqrt{1-x}$ and $g(x) = -x - \sqrt{1-x}$. Clearly these are defined only for $x\le 1$. Now f'(x) = -1 - \frac{1}{2\sqrt{1-x}} , so f'(x)<-1 for all $x<1$; this implies that $f(x)$ attains its minimum at $x=1$, where $f(1) = -1$, and that $\lim\limits_{x\to -\infty}f(x) = \infty$. Thus, the set of possible values of $m$ includes at least $[1,\infty)$.
g'(x) = -1 + \frac{1}{2\sqrt{1-x}} = 0 when $\sqrt{1-x} = \frac12$, i.e., when $x = \frac34$, and it’s easy to check that $g(x)$ has a minimum of $-\frac54$ at this point. Thus, the set of possible values of $m$ is actually $\left[-\frac54,\infty\right)$.