So, assuming $M$ is a pure (càdlàg) jump martingale, we get $$[M,M]_t = \sum_{ 0\leq s \leq t} (\Delta M_s)^2.$$ Of course the continous part $[M,M]_t^c$ of $[M,M]_t$ is exactly zero. That means by Meyer-Itô we get
$$M_t^2 = M_0^2 +\int_0^t 2 M_{s-}^2 dM_s+\sum_{ 0\leq s \leq t} \left[ M_s^2 - M_{s-}^2 - 2 M_{s-}\Delta M_s \right] $$
So because $M$ is a pure jump martingale we get $$\sum_{ 0\leq s \leq t} \left[ M_s^2 - M_{s-}^2 \right]= M_t^2 - M_0^2,$$ right? But with $$ \int_0^t 2 M_{s-}^2 dM_s = \sum_{ 0\leq s \leq t}2 M_{s-}\Delta M_s$$ we would get
$$M_t^2 = M_t^2$$
Okay, this is true... But I hoped to use that for showing that the compenssator of $[M,M]$ is teh same as the one of $M^2$...
EDIT: So I just solved it... Seems like I was confused :-)
So define the preservable quadratic variation $\langle M, M \rangle $ of $M$ as the unique adapted preservable process of finite variation with non-decreasing paths for which $M^2- \langle M, M \rangle$ is a (local) martingale (existence and uniqueness is secured by the Doob-Meyer-Decomposition.
So with $$\left( \Delta M_s \right)^2 =\Delta \left( M_s^2 \right)- 2M_{ s^- }\Delta M_s $$ and Itô we get $$ M_t^2- M_0^2 = 2 \int_0^t M_{s-} d{M_s} + \sum_{ i: \tau_i \leq t} \left[ M_{\tau_i}^2 - M_{\tau_i - }^2 - 2 M_{\tau_i -} \Delta M_{\tau_i} \right] \\=2 \int_0^t M_{s-} d{M_s} + \sum_{ i: \tau_i \leq t} \left( \Delta M_{\tau_i} \right)^2 =\int_0^t M_{s-} d{M_s} + [M,M]_t. $$ which implies $$[M,M]_t= M_t^2- M_0^2-2 \int_0^t M_{s-} d{M_s}$$ and so the compensator of $[M,M]$ must be the one of $M_t^2- M_0^2$ because the one of $\int_0^t M_{s-} d{M_s}$ is zero (it is a local martingale and following Doob-Meyer they are unique).
So it is possible to define $ \langle M, M \rangle$ as the unique adapted preservable process of finite variation with non-decreasing paths for which $$\left[ M,M \right]- \langle M, M \rangle$$ is a (local) martingale.
And the fun fact is: If we have a martingale which is not a pure jump martingale, we would get $$ M_t^2- M_0^2 = 2 \int_0^t M_{s-} d{M_s}+ [M,M]_t^c + \sum_{ i: \tau_i \leq t} \left[ M_{\tau_i}^2 - M_{\tau_i - }^2 - 2 M_{\tau_i -} \Delta M_{\tau_i} \right] \\=2 \int_0^t M_{s-} d{M_s} + + [M,M]_t^c+ \sum_{ i: \tau_i \leq t} \left( \Delta M_{\tau_i} \right)^2 =\int_0^t M_{s-} d{M_s} + [M,M]_t. $$ which would give us the same result!