Determine stability of feedback system from open loop transfer function and Nyquist stability criterion gives different results
However when I look at the closed loop transfer function, I would say that this system is unstable for =−1. In this case the transfer function becomes infinity so a bounded input will result in a unbounded (=infinity) output.

This depends on your definition of stability.  $GH = -1$ is called marginally stable because depending on how you look at it, it could be stable, or it could be unstable.
In the Lyapunov sense that @Petrus1904 mentions, it is stable.  But if you take bounded-input bounded-output (BIBO) stability to mean that the input can go on infinitely long but within bounds, then the output can, indeed, go to infinity.  So in the BIBO sense it's unstable.
Hence, &quot;marginally stable&quot;.

In my train of thought the point =−2 would again be stable since  will be finite again, however conform the Nyquist stability criterion this point will still be unstable?

An unstable system can have a transfer function that never goes infinite for $s \in j \omega$.  All it requires is that there are no poles on the stability boundary.
I apologize for not taking the time to do so, but I know that I could dummy up a system that has $GH = -2$ and was still stable, because I've designed such systems.  All you need to do is wrap a double integrator with a PID controller: let $G = 1/s^2$ and $H = k_i / s + k_p + \frac{s}{\tau_d s + 1} k_d$ and tune for stability.  The resulting system will have a spot in its open-loop Bode plot which has a phase shift of 180$^\circ$ and a gain greater than one -- to get that gain equal to two you'll just need to jigger the numbers around.
Addendum:  Use the PID transfer function I give above with $k_i = 0.002$, $k_p = 0.02$, $k_d = 0.2$ and $\tau_d = 0.01$ and you should get a system that's stable, with $GH \simeq -2$ at somewhere around $\omega = 0.1 \mathrm{\frac{rad}{sec}}$.
Assuming I got my math right...