Let $p,q$ and $s{}$ be prime numbers such that $2^sq =p^y-1$ where $y > 1.$ Find all possible values of $p.$
1. **Applying Zsigmondy's Theorem:**
   Zsigmondy's theorem states that for any integers \(a > b > 0\) and \(n > 1\), there exists a prime number that divides \(a^n - b^n\) and does not divide \(a^k - b^k\) for any \(k < n\), with some exceptions. In our case, \(a = p\) and \(b = 1\), so \(p^y - 1\) has a primitive prime divisor unless \(y = 2\) or \((p, y) = (2, 6)\).

2. **Case \(y > 4\):**
   If \(y > 4\) and \(y\) is not prime, then \(p^y - 1\) has two primitive prime divisors by Zsigmondy's theorem, which contradicts the given equation \(2^s q = p^y - 1\). Therefore, \(y\) must be a prime number or \(y = 4\).

3. **Case \(y = 4\):**
   If \(y = 4\), then \(p^4 - 1 = (p^2 - 1)(p^2 + 1)\). For \(p^4 - 1\) to be a power of two times a prime, \(p^2 - 1\) must be a power of two. The only prime \(p\) for which \(p^2 - 1\) is a power of two is \(p = 3\), since \(3^2 - 1 = 8 = 2^3\). However, \(3^4 - 1 = 80 = 2^4 \cdot 5\), and \(s = 4\) is not prime. Thus, \(p = 3\) is not a solution in this case.

4. **Case \(y \ge 3\) is prime:**
   Using the Lifting The Exponent (LTE) lemma, we have:
   \[
   v_2(p^y - 1) = v_2(p - 1) + v_2(y)
   \]
   Since \(2^s q = p^y - 1\), we get \(v_2(p^y - 1) = s\). Therefore, \(p - 1 = 2^s\). This implies \(p = 2^s + 1\). For \(p\) to be prime, \(s\) must be a power of 2, say \(s = 2^t\). The smallest such \(s\) is \(s = 2\), giving \(p = 5\). Indeed, \(5^3 - 1 = 124 = 2^2 \cdot 31\), which is a valid solution.

5. **Case \(y = 2\):**
   If \(y = 2\), then \(p^2 - 1 = 2^s q\). This can be factored as \((p - 1)(p + 1) = 2^s q\). Since \(p - 1\) and \(p + 1\) are consecutive even numbers, one must be a power of 2. We consider two sub-cases:
   - If \(p - 1 = 2^{s-1}\) and \(p + 1 = 2q\), then \(2q = 2^{s-1} + 2\), so \(q = 2^{s-2} + 1\) and \(p = 2^{s-1} + 1\). For \(q\) to be prime, \(s - 2\) must be a power of 2. The smallest such \(s\) is \(s = 3\), giving \(p = 5\).
   - If \(p = 2^{s-1} - 1\), then \(s - 1\) must be prime. The smallest such \(s\) is \(s = 3\), giving \(p = 3\). Indeed, \(3^2 - 1 = 8 = 2^3\).

Hence, the two possible values of \(p\) are \(3\) and \(5\).

The final answer is \(\boxed{3 \text{ and } 5}\)