Given the sequence $\{y_n\}_{n=1}^{\infty}$ defined by $y_1=y_2=1$ and
\[y_{n+2} = (4k-5)y_{n+1}-y_n+4-2k, \qquad n\ge1\]
find all integers $k$ such that every term of the sequence is a perfect square.
1. **Initial Terms and Recurrence Relation:**
   Given the sequence $\{y_n\}_{n=1}^{\infty}$ defined by $y_1 = y_2 = 1$ and the recurrence relation:
   \[
   y_{n+2} = (4k-5)y_{n+1} - y_n + 4 - 2k, \qquad n \ge 1
   \]

2. **Finding $y_3$:**
   Using the recurrence relation for $n=1$:
   \[
   y_3 = (4k-5)y_2 - y_1 + 4 - 2k = (4k-5) \cdot 1 - 1 + 4 - 2k = 4k - 5 - 1 + 4 - 2k = 2k - 2
   \]
   For $y_3$ to be a perfect square, $2k - 2$ must be a perfect square. Let $2k - 2 = m^2$ for some integer $m$. Then:
   \[
   2k = m^2 + 2 \implies k = \frac{m^2 + 2}{2}
   \]
   Since $k$ must be an integer, $m^2 + 2$ must be even, implying $m$ must be even. Let $m = 2a$ for some integer $a$. Then:
   \[
   k = \frac{(2a)^2 + 2}{2} = \frac{4a^2 + 2}{2} = 2a^2 + 1
   \]
   Therefore, $k$ must be of the form $2a^2 + 1$.

3. **Finding $y_5$:**
   Using the recurrence relation for $n=3$:
   \[
   y_4 = (4k-5)y_3 - y_2 + 4 - 2k = (4k-5)(2k-2) - 1 + 4 - 2k
   \]
   Substituting $k = 2a^2 + 1$:
   \[
   y_4 = (4(2a^2 + 1) - 5)(2(2a^2 + 1) - 2) - 1 + 4 - 2(2a^2 + 1)
   \]
   Simplifying:
   \[
   y_4 = (8a^2 - 1)(4a^2) - 1 + 4 - 4a^2 - 2 = 32a^4 - 4a^2 - 1 + 4 - 4a^2 - 2 = 32a^4 - 8a^2 + 1
   \]
   Using the recurrence relation for $n=4$:
   \[
   y_5 = (4k-5)y_4 - y_3 + 4 - 2k = (4(2a^2 + 1) - 5)(32a^4 - 8a^2 + 1) - (2(2a^2 + 1) - 2) + 4 - 2(2a^2 + 1)
   \]
   Simplifying:
   \[
   y_5 = (8a^2 - 1)(32a^4 - 8a^2 + 1) - 2a^2 + 2 + 4 - 4a^2 - 2 = 256a^6 - 96a^4 + 8a^2 + 1
   \]

4. **Checking if $y_5$ is a perfect square:**
   We need to check if $256a^6 - 96a^4 + 8a^2 + 1$ is a perfect square. We can use the inequality:
   \[
   (16a^3 - 3a^2 - 1)^2 < 256a^6 - 96a^4 + 8a^2 + 1 \leq (16a^3 - 3a^2)^2
   \]
   This inequality holds for $a = 1$, giving $k = 3$.

5. **Verifying $k = 3$:**
   For $k = 3$, the recurrence relation becomes:
   \[
   y_{n+2} = 7y_{n+1} - y_n - 2
   \]
   Let $f_n$ be the $n$-th Fibonacci term. From Catalan's Identity, we have:
   \[
   f_{n+2}f_n = (-1)^{n-1} + f_n^2
   \]
   and
   \[
   f_{n+2} = 3f_n - f_{n-2}
   \]
   For all odd $n$, we have:
   \[
   f_{n+2}^2 = 7f_n^2 - f_{n-2}^2 - 2
   \]
   By induction, $y_n = f_{2n-3}^2$ for all $n \ge 2$.

Conclusion:
\[
\boxed{k = 3}
\]