I am working on the solution of the following SDE:
$\begin{cases}\mathrm{d}Z_t = B_tZ_t \mathrm{d}t + B_tZ_t\mathrm{d}B_t\\ Z_0=1 \end{cases}$
Here $(Z_t)_{t\geq0}$ is the unknown process, while $(B_t)_{t\geq 0}$ is a (known) Brownian motion.
First, I tried to solve the homogeneous equation: $$ \mathrm{d}Z_t = B_tZ_t\mathrm{d}t$$
By Ito's formula, we have:
$$\mathrm{d}(\log Z_t) = \frac{1}{Z_t}\mathrm{d}Z_t - \frac{1}{2Z_t^2} (\mathrm{d}Z_t)(\mathrm{d}Z_t)$$ Substituting the homogenous equation, noticing that the term $(\mathrm{d}Z_t)(\mathrm{d}Z_t)$ vanishes, we get: $$\mathrm{d}(\log Z_t) = B_t\mathrm{d}t$$ Integration then yields: $$\log Z_t - \log Z_0 = \int_0^t B_s \mathrm{d}s$$ Finally, by the deterministic initial condition $Z_0 = 1$, we have $Z_t = \exp\left\{\int_0^t B_s\mathrm{d}s\right\} $
We then move on to attempting to solve the original equation by variation of constants. Let $(C_t)_{t\geq 0}$ be a (yet unknown) Ito process and $V_t = C_t Z_t$ with the result of $Z_t$ above. We want $\mathrm{d}V_t = B_tV_t \mathrm{d}t + B_tV_t\mathrm{d}B_t$.
By the product rule for Ito processes we have: $$\mathrm{d}V_t = Z_t \mathrm{d}C_t + C_t \mathrm{d}Z_t + (\mathrm{d}C_t)(\mathrm{d}Z_t)=$$ $$ = Z_t \mathrm{d}C_t + C_t B_t Z_t\mathrm{d}t + (\mathrm{d}C_t)(B_tZ_t\mathrm{d}t)=$$ $$ = Z_t \mathrm{d}C_t + B_t V_t \mathrm{d}t + B_tZ_t (\mathrm{d}C_t)(\mathrm{d}t) $$ where the last term vanishes, as the function $t \mapsto t$ is smooth.
Thus, what we want is the following:
$$ Z_t \mathrm{d}C_t = B_tV_t \mathrm{d}B_t \iff \frac{1}{C_t}\mathrm{d}C_t = B_t \mathrm{d}B_t $$
The problem with this is that by taking the integral, we get $$ \int_0^t \frac{1}{C_s} \mathrm{d}C_s = \int_0^t B_s \mathrm{d}B_s$$ and while the RHS is known, we can't really say anything about $C_t$, as de stochastic differential of the logarithm of $C_t$ would be $\mathrm{d}(\log C_t) = \frac{1}{C_t}\mathrm{d}C_t - \frac{1}{2C_t^2}(\mathrm{d}C_t)(\mathrm{d}C_t)$.
Any thoughts on how I could finish this solution?
You are already on the right track. To solve for $C_t$, do exactly like for $Z_t$. Indeed, from $$\frac{dC_t}{C_t} = B_tdB_t$$ you compute $d(\ln(C_t)$ by Ito's formula $$\begin{align} d(\ln(C_t)) &= \frac{dC_t}{C_t} -\frac{1}{2}\left(\frac{dC_t}{C_t},\frac{dC_t}{C_t} \right)\\ &= B_tdB_t -\frac{1}{2}\left(B_tdB_t,B_tdB_t \right)\\ &=B_tdB_t -\frac{1}{2}B_t^2\left(dB_t,dB_t \right)\\ &=B_tdB_t -\frac{1}{2}B_t^2dt\\ \end{align}$$ You deduce $$\ln(C_t)=\ln(C_0)-\frac{1}{2}\int_0^tB_s^2ds+\int_0^tB_sdB_s$$ and also the solution $V_t=C_tZ_t$.