Let $k$ be a natural number such that $k\ge 7$. How many $(x,y)$ such that $0\le x,y<2^k$ satisfy the equation $73^{73^x}\equiv 9^{9^y} \pmod {2^k}$?

[i]Proposed by Mahyar Sefidgaran[/i]
1. **Understanding the Problem:**
   We need to find the number of pairs \((x, y)\) such that \(0 \leq x, y < 2^k\) and \(73^{73^x} \equiv 9^{9^y} \pmod{2^k}\) for \(k \geq 7\).

2. **Using Lifting The Exponent Lemma (LTE):**
   Let \(v_2(n)\) denote the exponent of 2 in \(n\). The LTE lemma helps in finding \(v_2(a^b - c^d)\) under certain conditions. 

3. **Simplifying the Problem:**
   We need to consider the set \(S\) of all \(73^{73^x}\) for \(0 \leq x < 2^k\). We know that \(73 \equiv 9 \pmod{2^7}\) because \(73 = 9 + 2^7\).

4. **Lemma 1:**
   For any non-negative even integer \(a\), \(73^{73^a} \equiv 9^{9^a} \equiv 73 \equiv 9 \pmod{2^7}\).
   - Proof: Using LTE, \(v_2(73^{73^a} - 73) = v_2(73 - 1) + v_2(73^a - 1) = 3 + 3 + v_2(a) \geq 7\). Similarly for \(9^{9^a}\).

5. **Lemma 2:**
   For each odd \(y\) such that \(0 \leq y < 2^k\), there are exactly \(2^6\) numbers \(x\) with \(0 \leq x < 2^k\) such that \(73^{73^{x^8 + x}} \equiv 73^{73^{y^8 + y}} \pmod{2^k}\).
   - Proof: For odd \(x, y\), \(73^{73^{x^8 + x}} \equiv 73^{73^{y^8 + y}} \pmod{2^k}\) if and only if \(2^k \mid 73^{73^{x^8 + x} - 73^{y^8 + y}} - 1\). Using LTE, \(v_2(73^{73^{x^8 + x} - 73^{y^8 + y}} - 1) = 6 + v_2(x - y)\). Thus, \(k - 6 \leq v_2(x - y)\) implies \(2^{k-6} \mid x - y\).

6. **Counting the Solutions:**
   The interval \([0, 2^k - 1]\) is split into \(2^6\) intervals, each of length \(2^{k-6}\). For each odd \(y\), there are exactly \(2^6\) numbers \(x\) such that \(2^{k-6} \mid x - y\).

7. **Combining Results:**
   The set \(S\) has \(2^{k-1}\) elements, divided into \(2^6\) subsets \(S_i\) each of size \(2^{k-7}\). Each \(S_i\) contains elements congruent to \(9 \pmod{2^7}\). For each odd \(y\), there is exactly one element in \(S_i\) congruent to \(9^{9^{y^8 + y}} \pmod{2^k}\).

8. **Final Calculation:**
   The number of solutions is \(2^6 \times 2^{k-1} = 2^{k+5}\).

The final answer is \(\boxed{2^{k+5}}\).