Yes, $Z$ is a proper martingale. However, $\int_0^T(Z_sW_s)^2\,ds$ is not integrable for large $T$. As the quadratic variation of $Z$ is $[Z]_t=4\int_0^t(Z_sW_s)^2\,ds$, Ito's isometry says that this is integrable if and only if $Z$ is a square-integrable martingale, and you can show that $Z$ is not square integrable at large times (see below).
However, it is conditionally square integrable over small time intervals.
$$
\begin{align}
\mathbb{E}\left[Z_t^2W_t^2\;\Big\vert\;\mathcal{F}_s\right]&\le\mathbb{E}\left[W_t^2\exp(W_t^2)\;\Big\vert\;\mathcal{F}_s\right]\\
&=\frac{1}{\sqrt{2\pi(t-s)}}\int x^2\exp\left(x^2-\frac{(x-W_s)^2}{2(t-s)}\right)\,dx
\end{align}
$$
It's a bit messy, but you can evaluate this integral and check that it is finite for $s \le t < s+\frac12$. In fact, integrating over the range $[s,s+h]$ (any $h < 1/2$) with respect to $t$ is finite. So, conditional on $W_s$, you can say that $Z$ is a square integrable martingale over $[s,s+h]$.
This is enough to conclude that $Z$ is a proper martingale. We have $\mathbb{E}[Z_t\vert\mathcal{F}_s]=Z_s$ (almost surely) for any $s \le t < s+\frac12$. By induction, using the tower rule for conditional expectations, this extends to all $s < t$. Then, $\mathbb{E}[Z_t]=\mathbb{E}[Z_0] < \infty$, so $Z$ is integrable and the martingale conditions are met.
I mentioned above that the suggested method in the question cannot work because $Z$ is not square integrable. I'll elaborate on that now. If you write out the expected value of an expression of the form $\exp(aX^2+bX+c)$ (for $X$ normal) as an integral, it can be seen that it becomes infinite exactly when $a{\rm Var}(X)\ge1/2$ (because the integrand is bounded away from zero at either plus or minus infinity). Let's apply this to the given expession for $Z$.
The expression for $Z$ can be made more manageable by breaking the exponent into independent normals. Fixing a positive time $t$, then $B_s=\frac{s}{t}W_t-W_s$ is a Brownian bridge independent of $W_t$. Rearrange the expression for $Z$
$$
\begin{align}
Z_t&=\exp\left(W_t^2-\int_0^t(2(\frac{s}{t}W_t+B_s)^2+1)\,ds\right)\\
&=\exp\left(W_t^2-2\int_0^t\frac{s^2}{t^2}W_t\,ds+\cdots\right)\\
&=\exp\left((1-2t/3)W_t^2+\cdots\right)
\end{align}
$$
where '$\cdots$' refers to terms which are at most linear in $W_t$. Then, for any $p > 0$,
$$
Z_t^p=\exp\left(p(1-2t/3)W_t^2+\cdots\right).
$$
The expectation $\mathbb{E}[Z_t^p\mid B]$ of $Z_t^p$ conditional on $B$ is infinite whenever
$$
p(1-2t/3){\rm Var}(W_t)=p(1-2t/3)t \ge \frac12.
$$
The left hand side of this inequality is maximized at $t=\frac34$, where it takes the value $3p/8$. So, $\mathbb{E}[Z_{3/4}^p\mid B]=\infty$ for all $p\ge\frac43$. The expected value of this must then be infinite, so $\mathbb{E}[Z^p_{3/4}]=\infty$. It is a standard application of Jensen's inequality that $\mathbb{E}[\vert Z_t\vert^p]$ is increasing in time for any $p\ge1$ and martingale $Z$. So, $\mathbb{E}[Z_t^p]=\infty$ for all $p\ge 4/3$ and $t\ge3/4$. In particular, taking $p=2$ shows that $Z$ is not square integrable.
Define $X_t = -\int_0^t\beta_s\,dW_s$ and $Y_t = -\frac{1}{2}\int_0^t\beta_s^2\,ds$. Then $Z_t = e^{X_t+Y_t}$. Even though you did not mention it I am guessing that there is a condition on $(\beta_s)_{s\geq 0}$ that makes $X$ a local martingale. Since Brownian motion has continuous paths, $X$ is continuous as well, which implies $Z$ is continuous as well. Hence, (needless to say) $X$ and $Y$ are both semimartingales. Since $(x,y) \mapsto e^{x+y}$ is twice continuously differentiable in all its arguments, Ito's formula applies, i.e.
$$dZ_t = Z_tdX_t + Z_tdY_t + \frac{1}{2}Z_td[X,X]_t+ \frac{1}{2}Z_td[Y,Y]_t + Z_td[X,Y]_t$$
You can check the following identities yourself.
$$dX_t = -\beta_tdW_t$$
$$dY_t = -\frac{1}{2}\beta_t^2dt$$
$$d[X,X]_t = \beta^2_tdt$$
$$d[X,Y]_t = d[Y,Y]_t = 0$$
Substituting these identities into the SDE above,
$$dZ_t = -Z_t\beta_tdW_t - Z_t\frac{1}{2}\beta_t^2dt + Z_t\frac{1}{2}\beta^2_tdt$$
Hence, $dZ_t = -Z_t\beta_tdW_t$. I mentioned above that there must be a condition on $\beta$ that makes $X$ a local martingale. The most general condition for this is that
$$\int_0^t\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\triangle)$$
for every $t < \infty$. We will require the same for $Z$ to be a local martingale, i.e.
$$\int_0^tZ_s^2\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\square)$$
for every $t < \infty$. Now fix an arbitrary $0 < t < \infty$. Let $\Omega_t$ denote the full measure set where $(\triangle)$ is true. Let $\Omega_c$ be the full measure set where $Z$ is continuous. Fix $\omega \in \Omega_c \cap \Omega_t$. Since $s \mapsto Z_s(\omega)$ is continuous, $s \mapsto 1_{[0,t]}(s)Z_s(\omega)$ is bounded. Then,
$$\int_0^tZ_s(\omega)^2\beta_s(\omega)^2\,ds \leq \sup_{s \in [0,t]}Z_s(\omega)^2 \int_0^t\beta_s(\omega)^2\,ds < \infty$$
Since $\Omega_c \cap \Omega_t$ has full measure, $(\square)$ is true. So then, $Z$ is a local martingale.
Best Answer
A possible approach here would be to use some basic Itô calculus.
If we compute the stochastic differential of $X_t$ using Itô's chain rule , we get $$ d X_t = X_t \left( -\frac 12 \sigma_t^2 dt + \sigma_t dB_t \right) + \frac 12 X_t \sigma_t^2 dt = X_t \sigma_t dB_t. $$ Hence the differential of $X_t$ has no drift, implying that $X_t$ is an Itô integral, i.e. $$ X_t = \int_0^t X_s \sigma_s d B_s , $$ which is well-known to be a martingale.