Consider the many shared resource scheduling problem where jobs have to be scheduled on identical parallel machines with the goal of minimizing the makespan. However, each job needs exactly one additional shared resource in order to be executed and hence prevents the execution of jobs that need the same resource while being processed. Previously a $(2m/(m+1))$-approximation was the best known result for this problem. Furthermore, a $6/5$-approximation for the case with only two machines was known as well as a PTAS for the case with a constant number of machines. We present a simple and fast 5/3-approximation and a much more involved but still reasonable 1.5-approximation. Furthermore, we provide a PTAS for the case with only a constant number of machines, which is arguably simpler and faster than the previously known one, as well as a PTAS with resource augmentation for the general case. The approximation schemes make use of the N-fold integer programming machinery, which has found more and more applications in the field of scheduling recently. It is plausible that the latter results can be improved and extended to more general cases. Lastly, we give a $5/4 - \varepsilon$ inapproximability result for the natural problem extension where each job may need up to a constant number (in particular $3$) of different resources.