To further elaborate on my comment, it is a theorem that if $X^1,X^2,\ldots,X^n$ are Lévy processes with respect to a common filtration, all starting from zero, then they are independent *if and only if* their quadratic covariations $[X^i,X^j]$ are all (almost surely) zero. This is stated as Theorem 11.43 of He, Wang & Yan, *Semimartingale Theory and Stochastic Calculus*. It's not difficult to prove with a bit of stochastic calculus, and I'll give a proof for two Lévy processes below.

In the situation described in the question, there are two Lévy processes $W$ and $N$, where $N$ is a pure jump process. So, the quadratic covariation is simply a sum over the jumps of the processes.
$$
[W,N]\_t=\sum_{s\le t}\Delta W_s\Delta N_s.
$$
But as Brownian motion is continuous, $\Delta W$ is zero, so the covariation $[W,N]$ is zero. Therefore, they are independent.

Now, let's show that if $X$, $Y$ are Lévy processes w.r.t. the filtration $\{\mathcal{F}\_t\}\_{t\in\mathbb{R}^+}$ with $X_0=Y_0=0$ and $[X,Y]=0$ then they are independent. The characteristic functions of $X$ and $Y$ can be written as
$$
\begin{align}
&\mathbb{E}\left[e^{iaX_t}\right]=\exp(t\psi_X(a)),\\\\
&\mathbb{E}\left[e^{iaY_t}\right]=\exp(t\psi_Y(a)).
\end{align}
$$
Independence of the increments w.r.t. $\mathcal{F}\_{\cdot}$ implies that $M_t\equiv\exp(iaX_t-t\psi_X(a))$ and $N_t\equiv\exp(ibY_t-t\psi_Y(b))$ are martingales. As the jumps of the quadratic covariation equals the product of the jumps of the processes, $\Delta [X,Y]=\Delta X\Delta Y$, it follows that $X$ and $Y$ cannot jump simultaneously. So, $\Delta [M,N]=\Delta M\Delta N=0$. Also, the continuous part of the quadratic covariation $[M,N]^{c}$ is just an integral with respect to $[X,Y]^{c}$ (which follows from Ito's formula for non-continuous semimartingales). So, the covariation $[M,N]$ is zero. Using integration by parts,
$$
d(M_tN_t)=M_{t-}dN_t + N_{t-}dM_t+d[M,N]\_t=M_{t-}dN_t + N_{t-}dM_t.
$$
As a sum of integrals with respect to martingales, $MN$ is a local martingale. As it is also bounded at any time, this is a proper martingale. So, $\mathbb{E}[M_tN_t\mid\mathcal{F}\_s]=M_sN_s$ for $s < t$. Plugging in the definitions of $M$ and $N$,
$$
\mathbb{E}\left[e^{iaX_t+bY_t}\;\Big\vert\;\mathcal{F}\_s\right]=\exp(iaX_s+(t-s)\psi_X(a))\exp(ibY_s+(t-s)\psi_Y(b)).
$$
This determines the joint characteristic function of $(X_t,Y_t)$ conditional on $\mathcal{F}\_s$, showing that they are independent. As the distributions of $(X_t,Y_t)$ conditional on $\mathcal{F}\_s$ ($s < t$) determine all finite distributions, $X$ and $Y$ are independent. I'll leave the converse ($X,Y$ independent implies $[X,Y]=0$) as an exercise. It's not needed for the question anyway.

You can also compare this with the argument given by kakuritsu. It is essentially the same thing. Rather than working under the generality of Lévy processes, he (or she?) works directly with the Brownian motion and Poisson process, for which $\psi_W(a)=-\frac12a^2$ and $\psi_N(a)=\lambda(e^{ia}-1)$, and uses the moment generating rather than characteristic function (effectively, $a$ and $b$ above are imaginary).