How to deal with $\forall M > 0$ in this limit?

For the limit $$\lim_{x\to +\infty} x^2+3 = +\infty$$ by definition we have that $\forall M > 0, \exists N > 0$ such that $\forall x > N$ we have $f(x) > M$.

But in this case then $\forall x > N$ we have $x^2 + 3 > M$ that is $x^2 > M - 3$ that is again $x > \sqrt{M - 3}$ and $x < -\sqrt{M-3}$. Considering the positive case, we have

$$x > \sqrt{M - 3}$$

But this is not true $\forall M > 0$, since for $M = 0$ or $M = 1$ or $M = 8/19$ the square root doesn't exist.

Then it should be $N_M = \sqrt{M-3}$ but things stopped working before this...

How should I reason with these types of limits/proofs?