This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations":
Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the stochastic integral $H\cdot M$ is a local martingale.
I'm reading that book, but haven't got that far yet, will be interested to see if someone can provide a simple proof just for this special case in this post.
After going on a literature hunt I found an answer in this script on page 30.
The integral is decomposed into its off-diagonal terms (first row) and its diagonal/quadratic terms (second row).
The scaling factor of 2 occurs since we're integrating twice over the same domain.
Now to the quadratic term:
We are working with a compensated Poisson process which means that the process is centered around zero.
$$
\mathbb{E}[Y_t] = \lambda \mathbb{E}[Z] t
$$
Discretizing the integral gives us:
$$
\sum (\Delta \phi_{t-})^2 \Delta[Y_t - \lambda \mathbb{E}[Z]]\ \Delta[Y_t - \lambda \mathbb{E}[Z]]
$$
Substituting in the mean $\mathbb{E}[Y_t] = \mathbb{E}[Y_t]$ yields:
$$
\sum_t (\Delta \phi_{t-})^2 \Delta[Y_t - \mathbb{E}[Y_t]]^2
$$
where $ \Delta[Y_t - \mathbb{E}[Y_t]]^2$ is the diagonal term of the covariance for an infinitessimal small timestep $\Delta$.
We further know that the variance for a Poisson process $N_t$ of length $t$ is
$$
\mathbb{V}[N_t] = \lambda t
$$
and that the variance of a scaled random variable $X$ is
$$
\mathbb{V}[aX] = a^2 \mathbb{V}[X].
$$
Since the jumps of the compensated Poisson process are modulated of sorts by the random variable $Z$ we obtain:
$$
\sum_t (\Delta \phi_{t-})^2 \Delta[Y_t - \mathbb{E}[Y_t]]^2 \\
= \sum_t (\Delta \phi_{t-})^2 \mathbb{V}[Z \Delta N_t] \\
= \sum_t (\Delta \phi_{t-})^2 Z^2 \mathbb{V}[\Delta N_t] \\
= \sum_t (\Delta \phi_{t-})^2 Z^2 \lambda \Delta t
$$
the continuous version of which is
$$
\int \phi_{t-}^2 Z^2 \lambda dt
$$
Best Answer
You can see this book Introduction to Stochastic Integration p.109 and following; example 7.6.3 answers your question.
Sorry for my english.