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.
As @Did points out, by definition of $Y_t$ in terms of stochastic integrals,
$$
\text dY_t = \frac{1}{X_t}\text dX_t-\frac{1}{2}\frac{1}{X_t^2}\text d\langle X\rangle_t\tag{1}
$$
And
$$
Z_t^{-1}\text dZ_t = Z_t^{-1}\text de^{Y_t}=Z_t^{-1}e^{Y_t}(\text dY_t+\frac{1}{2}\text d\langle Y\rangle_t) = \text dY_t+\frac{1}{2}\text d\langle Y\rangle_t\tag{2}
$$
Using (1) in (2), conclude.
Best Answer
The issue with your solution is that $Y_t Z_t = 1$ does not imply that $\langle Y_t, Z_t \rangle = 0$ for the covariation. See the answer by @ChristopherK for a computation of $\langle Y_t, Z_t \rangle$. Here's a way to solve your original problem.
Let $$R_t = \underbrace{\int_0^t X_u dW_u}_{(\star )} - \underbrace{\frac{1}{2}\int_0^t X_u^2 du}_{(\star \star)}$$ Note that with the assumption that $\int_0^t X_u^2 du < \infty$, the process $R_t$ is well-defined. Moreover, $R_t$ is a semi-martingale, because $(\star)$ is a local martingale and $(\star \star)$ is a process of bounded variation.
Let $g(x) = e^{-x}$ and observe that $Y_t = g(R_t)$. Itô's lemma implies that the class of semi-martingales is stable under the application of $C^2$ maps, which $g$ clearly is. That solves (1).
To solve (2), observe that $g_x(x) = - e^{-x}, g_{xx}(x) = e^x$. Further observe that the dynamics of $R_t$ can be written as $$dR_t = X_t dW_t - \frac{1}{2} X_t^2 dt$$ By Itô's lemma: $$\begin{align*} dY_t &= g_x(R_t)(dR_t) + \frac{1}{2} g_{xx}(R_t)(dR_t)^2 \\ &= - \exp (-R_t) \left( X_t dW_t - \frac{1}{2}X_t^2 dt\right) + \frac{1}{2} \exp(-R_t) X_t^2 dt \\ &=\exp (-R_t) X_t^2 dt - \exp (-R_t) X_t dW_t \\ &=Y_t X_t^2 dt - Y_tX_tdW_t \end{align*}$$
As required.
As an additional technical point, setting $Y_t = \frac{1}{Z_t}$ is a perfectly valid thing to do you because $P \left(\inf_{0 \leq s \leq t} Z_s > 0\right) = 1 $.