Play with integers

If only x is prime, then only y will be an integer. And as we know, there are infinite primes, so the answer is infinite. From Fermat's Little theorem, $a^{p-1}$ mod $p ≡ 1$ mod $p$ , putting $a = 2$ , $2^{p-1} = 1 + k.p$ , and as, $a = 2$ , so p can't be equal to 2 .

So, $\Rightarrow k-1 = \frac{2^{p-1}-(1+ p)}{p}$ , and as, $k$ is an integer, $k - 1$ , will obviously be an integer, only if p is prime.

Now, we are defining a function $y(x) = \frac{2^{x-1}-(1+ x)}{x}$ , thus, y will be an positive integer, only if x is a prime. And there are infinite primes, so number of values of x is infinite.