If both $ x$ and $ y$ are both integers, how many pairs of solutions are there of the equation $ (x\minus{}8)(x\minus{}10) \equal{} 2^y?$

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ \text{more than 3}$
To solve the problem, we need to find all integer pairs \((x, y)\) that satisfy the equation \((x-8)(x-10) = 2^y\).

1. **Expand and simplify the equation:**
   \[
   (x-8)(x-10) = x^2 - 18x + 80
   \]
   Therefore, we need to find integer values of \(x\) such that \(x^2 - 18x + 80 = 2^y\).

2. **Analyze the quadratic equation:**
   We need to check if \(x^2 - 18x + 80\) can be a power of 2. Let's denote \(2^y = k\), where \(k\) is a power of 2. Thus, we have:
   \[
   x^2 - 18x + 80 = k
   \]
   Rearrange this to:
   \[
   x^2 - 18x + (80 - k) = 0
   \]
   This is a quadratic equation in \(x\). For \(x\) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \(\Delta\) is given by:
   \[
   \Delta = b^2 - 4ac = (-18)^2 - 4 \cdot 1 \cdot (80 - k) = 324 - 4(80 - k) = 324 - 320 + 4k = 4 + 4k
   \]
   For \(\Delta\) to be a perfect square, \(4 + 4k\) must be a perfect square. Let \(4 + 4k = m^2\) for some integer \(m\). Then:
   \[
   4(1 + k) = m^2 \implies 1 + k = \left(\frac{m}{2}\right)^2
   \]
   Since \(k = 2^y\), we have:
   \[
   1 + 2^y = \left(\frac{m}{2}\right)^2
   \]
   This implies that \(\left(\frac{m}{2}\right)^2\) must be an integer, so \(m\) must be even. Let \(m = 2n\), then:
   \[
   1 + 2^y = n^2
   \]

3. **Solve for \(y\):**
   We need to find \(y\) such that \(n^2 - 1 = 2^y\). Let's test small values of \(y\):
   - For \(y = 0\): \(n^2 - 1 = 1 - 1 = 0\) (not a valid solution)
   - For \(y = 1\): \(n^2 - 1 = 2 - 1 = 1 \implies n = \pm 1\)
   - For \(y = 2\): \(n^2 - 1 = 4 - 1 = 3\) (not a valid solution)
   - For \(y = 3\): \(n^2 - 1 = 8 - 1 = 7\) (not a valid solution)
   - For \(y = 4\): \(n^2 - 1 = 16 - 1 = 15\) (not a valid solution)
   - For \(y = 5\): \(n^2 - 1 = 32 - 1 = 31\) (not a valid solution)
   - For \(y = 6\): \(n^2 - 1 = 64 - 1 = 63\) (not a valid solution)
   - For \(y = 7\): \(n^2 - 1 = 128 - 1 = 127\) (not a valid solution)
   - For \(y = 8\): \(n^2 - 1 = 256 - 1 = 255\) (not a valid solution)
   - For \(y = 9\): \(n^2 - 1 = 512 - 1 = 511\) (not a valid solution)
   - For \(y = 10\): \(n^2 - 1 = 1024 - 1 = 1023\) (not a valid solution)
   - For \(y = 11\): \(n^2 - 1 = 2048 - 1 = 2047\) (not a valid solution)
   - For \(y = 12\): \(n^2 - 1 = 4096 - 1 = 4095\) (not a valid solution)
   - For \(y = 13\): \(n^2 - 1 = 8192 - 1 = 8191\) (not a valid solution)
   - For \(y = 14\): \(n^2 - 1 = 16384 - 1 = 16383\) (not a valid solution)
   - For \(y = 15\): \(n^2 - 1 = 32768 - 1 = 32767\) (not a valid solution)
   - For \(y = 16\): \(n^2 - 1 = 65536 - 1 = 65535\) (not a valid solution)

   The only valid solution is when \(y = 1\), giving \(n = \pm 1\).

4. **Find corresponding \(x\) values:**
   - For \(n = 1\):
     \[
     1 + 2^1 = 2 \implies x^2 - 18x + 80 = 2 \implies x^2 - 18x + 78 = 0
     \]
     The discriminant is:
     \[
     \Delta = 18^2 - 4 \cdot 78 = 324 - 312 = 12 \quad (\text{not a perfect square})
     \]
   - For \(n = -1\):
     \[
     1 + 2^1 = 2 \implies x^2 - 18x + 80 = 2 \implies x^2 - 18x + 78 = 0
     \]
     The discriminant is:
     \[
     \Delta = 18^2 - 4 \cdot 78 = 324 - 312 = 12 \quad (\text{not a perfect square})
     \]

Since no valid \(x\) values are found, there are no integer solutions to the equation.

The final answer is \(\boxed{0}\)