= Solution
Since the density <martingale> is <uniformly integrable>, $Z_t=\mathbb E_{\mathbb P}[Z_\infty\mid\mathcal F_t]$. Equivalence of measures, as stipulated, means $Z_\infty>0$ almost surely. Write $L=\int Z^{-1}\,dZ$ for the <stochastic logarithm>. The <Itô formula> gives $\log Z=L-\langle L\rangle/2$, and hence the <quadratic covariation> in the drift correction is
$$
C_t=\langle\log Z,X\rangle_t=\int_0^t Z_s^{-1}\,d\langle Z,X\rangle_s.
$$
It involves the local-<martingale> part of $\log Z$; its continuous finite-variation part has zero covariation.
With $Y=X-C$, the <Itô product rule> now gives an exact cancellation:
$$
d(ZY)=Y\,dZ+Z\,dX-Z\,dC+d\langle Z,X\rangle=Y\,dZ+Z\,dX.
$$
Thus $ZY$ is a P-<local martingale>. To transfer this conclusion rigorously, set $\tau_n=\inf\{t:|Y_t|\geq n\}$. The same computation for the stopped $Y$ gives
$$
d(Z_tY_{t\wedge\tau_n})=Y_{t\wedge\tau_n}\,dZ_t+\mathbf1_{\{t\leq\tau_n\}}Z_t\,dX_t.
$$
This is again a P-<local martingale>. Moreover its absolute value is at most $nZ_t$. <Uniform integrability> of $Z$ makes the family over bounded <stopping times> <uniformly integrable>, so the product is a true <martingale>. This is the <bounded-process density-product criterion>.
The <Bayes formula for conditional expectation> therefore gives, for $s\leq t$,
$$
\mathbb E_{\mathbb Q}[Y_{t\wedge\tau_n}\mid\mathcal F_s]
=\frac{\mathbb E_{\mathbb P}[Z_tY_{t\wedge\tau_n}\mid\mathcal F_s]}{Z_s}
=Y_{s\wedge\tau_n}.
$$
Continuity gives $\tau_n\uparrow\infty$, proving \b[$X-\langle\log Z,X\rangle$ is a Q-<local martingale>]. This proves the needed <Girsanov theorem> rather than invoking it.
For the Brownian-filtration conclusion, use this precise <Brownian martingale representation theorem>: every <continuous local martingale> in the usual augmentation of the natural Brownian filtration is an Itô integral with a predictable integrand locally square integrable in time. In particular
$$
Z_t=1+\int_0^tH_s\,dW_s,\qquad\int_0^tH_s^2\,ds<\infty\quad\text{a.s.}
$$
Define
$$
\boxed{\alpha_s=H_s/Z_s.}
$$
On each finite horizon a strictly positive continuous $Z$ has a positive pathwise minimum. Thus $\int_0^t\alpha_s^2\,ds<\infty$ almost surely, and
$$
\langle\log Z,W\rangle_t=\int_0^t\alpha_s\,ds.
$$
The already proved measure-change result makes $\widehat W=W-\int\alpha\,ds$ a continuous Q-<local martingale>. Its <quadratic variation> is $t$, unchanged by its finite-variation correction. The <Lévy characterization of Brownian motion> proves
$$
\boxed{\widehat W_t=W_t-\int_0^t\alpha_s\,ds\text{ is Q-Brownian motion}.}
$$
No additional exponential-integrability condition is needed, since the equivalent <uniformly integrable> density process is already given.
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