Set
$$f : x\mapsto \int_0^x\exp\left(-2\beta y-\frac 23y^3\right)\mathrm dy. $$
The point of $f$ is that $f(X)$ is a local martingale, possibly up to the explosion time $\tau$ of $X$ ($f$ is solution to $(\beta+x^2)\partial_xf+\frac12\Delta f=0$, in fact all such solutions can be written as $af+b$).

Notice that $f$ is increasing. Moreover, it is bounded above, since the exponential term goes to zero fast enough. Since the explosion of $X$ can only occur if $X$ diverges to $+\infty$ ($X_t$ is at least $x_0+\beta t+B_t$), $f(X)$ is in fact a local martingale for all times, setting by convention $f(X_t)=\sup f$ for all $t\geq\tau$.

As a local martingale bounded above, $f(X)$ converges almost surely, and given that $f$ is increasing and isn't bounded below, $X$ converges in $\mathbb R\cup\{+\infty\}$. We need to show that $\lim X_t$ cannot be finite with positive probability.

Notice that
$$ B_{t+1}-B_t = (X_{t+1}-X_t) - \int_t^{t+1}(\beta+X_s^2)\mathrm ds. $$
In particular, if $X$ converges to a finite limit $\ell\in\mathbb R$, then $B_{t+1}-B_t$ converges to $ -\beta-\ell^2$. Hence the event that $X$ converges to a finite limit is included in the event that $B_{t+1}-B_t$ converges, which obviously has measure zero.