$\delta[n+1]$ equals to one only $n+1 = 0$ ?
If the 1. is true, so what does a signal $x[n+1] $ stand for? Is this signal causal or anticausal? I mean, if 1. is true, $n = -1$, so is this signal causal? (because it just happens before the time I want). But Oppenheim says that it is anticausal. Why?
A discrete signal can be defined at different integer time indices. Generally, one talks about causal or anticausal systems. But since a signal can, when convolved, be interpreted as a system, some say that:
Causal signals are signals that are zero for all negative time, while anticausal are signals that are zero for all positive time.
You can read that in the Classification of signals. Accordingly to that definition, your signal would be anticausal. Oppenheim sounds correct. One interpretation is: if you convolve a signal $x$ by an anticausal signal, the result will start "sooner" that $x$, a breach in causality.
The signal $x(n+1)$ just means that it is a signal based on $x(n)$ but with the samples at other time instances, in particular, all the samples are at the previous time instance. For example, the value that $x(n)$ had at $n=0$, now would be at $n=-1$, because to obtain $x(0)$ with the new signal $x(n+1)$ you have to put the time index $n=-1$, so that $x(-1+1) = x(0)$.
Based on the definition that was given in Laurent Duval's answer, if your signal had a non zero value at $n=0$ then after the shift it would become non causal.
Hope that helps.