Let us begin by showing that $X$ diverges to $+\infty$, possibly as time goes to infinity. 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.

So $X$ diverges to $+\infty$. I will now try to use deterministic arguments. Let us assume for now the following

**Fact.**

Let $Y$ be a process such that $$ Y_t \geq -C + \int_0^t \left(Y_s^2-\alpha^2\right)\mathrm ds $$
for some constants $C,\alpha>0$, and all $t$ possibly up to some explosion time $\tau$ (explosion means “leaves all compact subsets”).

Then either $\liminf_{t\to\tau} Y_t\leq\alpha$ or $Y$ diverges to $+\infty$ in finite time.

Note that for any $\alpha>0$ such that $\alpha^2>-\beta$, almost surely there exists a (random) constant $C>0$ such that $B_t\geq -C-(\alpha^2+\beta)t$ for all $t>0$. In particular,
$$ X_t \geq x_0 - C + \int_0^t\left(X_s^2 - \alpha^2\right)\mathrm ds. $$
Since $X$ diverges to $+\infty$, obviously its limit inferior is not bounded above, so the fact implies that $X$ must undergo explosion in finite time.

Now onto the proof of the fact. Suppose that the limit inferior of $Y$ is larger than $\alpha$. Then for all $t$ large enough (which here means close enough to $\tau$), $Y_t^2>\alpha^2+\varepsilon$ for some $\varepsilon>0$. According to the inequality, $Y$ must then diverge, possibly in infinite time.

Setting
$$ I_t = \int_0^t \left(Y_s^2-\alpha^2\right)\mathrm ds, $$
we see that
$$ I'_t = Y_t^2 - \alpha^2 \geq \frac12(Y_t+C)^2 \geq \frac12I_t^2 $$
for all $t$ large enough. In particular (note that $I_t>0$ for $t$ large enough),
$$ \frac{\mathrm d}{\mathrm dt}\left(-\frac1{I_t}\right)
= \frac{I'_t}{I_t^2}
\geq \frac12 $$
for all $t$ large enough, hence
$$ Y_t\geq -C + I_t\geq-C+\frac 2{T-t} $$
for some $T>0$ and all $t$ large enough, so that $Y$ explodes in finite time.