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.
This may not be exactly what you're looking for, but here's a proof which at least uses the specific form of the process (being the exponential local martingale of an integral of a square-integrable deterministic process with respect to a Brownian motion):
First off, in order to make things fit better with the standard framework, I'll assume that $\phi$ is a mapping from $[0,\infty)$ to $\mathbb{R}$ with the property that $\int_0^t \phi(s)^2ds$ is finite for all $t\ge0$. Define $N_t = \int_0^t \phi(s)dW_s$, we then have $M_t = \mathcal{E}(N)_t$, the exponential local martingale, and the objective is to prove that $\mathcal{E}(N)$ is a martingale (instead of just a local martingale). By some classical results, $\mathcal{E}(N)$ is a nonnegative supermartingale, and it is a martingale if and only if $E\mathcal{E}(N)_t = 1$ for all $t\ge0$. I don't know where you actually can find a proof of these claims, but they follow from applications of the optional sampling theorem and Fatou's lemma. The conclusion is that we need to show $E\mathcal{E}(N)_t=1$ for all $t\ge0$.
To do so, first note that $[N]_t = \int_0^t \phi(s)^2 ds$. We apply Itô's formula:
$\mathcal{E}(N)^2_t = 1 + 2\int_0^t \mathcal{E}(N)_sd\mathcal{E}(N)_s+[\mathcal{E}(N)]_t\\
=1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_sd[N]_s\\
=1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_s\phi(s)^2ds.$
As $\mathcal{E}(N)$ and $\mathcal{E}(N)\cdot \mathcal{E}(N)$ are continuous local martingales, they are locally bounded. Let $(T_n)$ be a localising sequence of stopping times such that both $\mathcal{E}(N)^{T_n}$ and $(\mathcal{E}(N)\cdot \mathcal{E}(N))^{T_n}$ are bounded martingales. By the martingale property of $(\mathcal{E}(N)\cdot\mathcal{E}(N))^{T_n}$, we then obtain
$E\mathcal{E}(N)_{t\land T_n}^2 = 1 + E\int_0^{t\land T_n}\mathcal{E}(N)_sd\mathcal{E}(N)_s
+ E\int_0^{t\land T_n}\mathcal{E}(N)^2_s\phi(s)^2 ds\\
=1 + E \int_0^t \mathcal{E}(N)^2_s\phi(s)^21_{[0,T_n]}(s)ds
\le 1 + \int_0^t E\mathcal{E}(N)^2_{s\land T_n}\phi(s)^2ds$
where we also applied Tonelli's theorem and nonnegativity of $\mathcal{E}(N)$. Now consider a fixed $n\ge1$ and define $g_n(t) = E\mathcal{E}(N)^2_{t\land T_n}$. The above then states that
$g_n(t) \le 1+ \int_0^t g_n(s)\phi(s)^2ds$.
By a classical analysis lemma, Gronwall's lemma (See the Wikipedia article on Gronwall's inequality), we then obtain $g_n(t) \le \exp(\int_0^t\phi(s)^2ds)$. In other words, we have now shown
$E\mathcal{E}(N)^2_{t\land T_n}\le \exp(\int_0^t \phi(s)^2ds)$
for all $n\ge1$ and $t\ge0$. Now fix $t\ge0$. By the above, the family $(\mathcal{E}(N)_{t\land T_n})_{n\ge1}$ is then bounded in $\mathcal{L}^2$, therefore uniformly integrable. Furthermore, $\mathcal{E}(N)_{t\land T_n}$ converges almost surely to $\mathcal{E}(N)_t$. Combining this with uniform integrability, $\mathcal{E}(N)_{t\land T_n}$ converges in $\mathcal{L}^1$ to $\mathcal{E}(N)_t$, and so the means also converge. And as $\mathcal{E}(N)^{T_n}$ is a martingale, $\mathcal{E}(N)_{t\land T_n}=1$. We conclude that $E\mathcal{E}(N)_t = 1$, and so $\mathcal{E}(N)$ is a martingale.
Best Answer
In order to make sense of a stochastic integral of the form $\int_0^t f(B_s) \, dB_s$ you need to verify two properties:
Under the above conditions, the stochastic integral $M_t := \int_0^t f(B_s) \, dB_s$ is a local martingale but, in general, it might fail to be a martingale. In order to ensure that $(M_t)_{t \geq 0}$ is a martingale (not only a local one), $f$ has to satisfy the stronger integrability condition of the above-mentioned conditions, that is, $\mathbb{E}(\int_0^t f(s)^2 \, ds)<\infty$ for all $t\geq 0$.
This is exactly the condition which the authors of the solution verified for the particular case $f(s):=B_s^2$.