Here is a general approach to find all square root of a $2\times2$ matrix $A$ which is not a multiple of the identity matrix.
Assume that $X^2 = A$. If we write $t = \operatorname{tr}(A)$ and $d = \det(A)$, then by the Cayley-Hamilton theorem we have
$$ X^2 - tX + dI = 0. $$
Plugging this to $X^2 = A$, we have $A = tX - dI$. Since $A$ is not a multiple of $I$, $t$ is automatically non-zero and we can write
$$X = \frac{1}{t}(A + dI). \tag{1}$$
So it suffices to determine $d$ and $t$. For $d$, we know that
$$d^2 = \det(X)^2 = \det(X^2) = \det(A) \qquad \Rightarrow \qquad d = \pm\sqrt{\det(A)}, \tag{2}$$
which can be computed. Similarly,
$$ \operatorname{tr}(A) = \operatorname{tr}(tX - dI) = t^2 - 2d \qquad \Rightarrow \qquad t = \pm \sqrt{2d + \operatorname{tr}(A)}. \tag{3}$$
This shows that if $X^2 = A$, then $X$ must be of the form $\text{(1)}$ with $d$ and $t$ as above. Conversely, if $d$ and $t$ are as above, then $X$ given by $\text{(1)}$ satisfies
\begin{align*}
X^2
&= \frac{1}{t^2}(A^2 + 2dA + d^2I) \\
&= \frac{1}{t^2}(\operatorname{tr}(A)A - \det(A)I + 2dA + d^2I) \\
&= \frac{1}{t^2}((t^2 - 2d)A - d^2I + 2dA + d^2I) \\
&= A.
\end{align*}
Therefore $X^2 = A$ if and only if $X$ is given by $\text{(1)}$ with $d$ and $t$ given by $\text{(2)}$ and $\text{(3)}$.
Now in your case,
$$ (A+B)^2 = \begin{pmatrix} 32 & 64 \\ 16 & 32 \end{pmatrix}
\quad \Rightarrow \quad d = 0
\quad \Rightarrow \quad t = \pm 8
\quad \Rightarrow \quad A+B = \pm \begin{pmatrix} 4 & 8 \\ 2 & 4 \end{pmatrix} $$
and likewise
$$ (A-B)^2 = \begin{pmatrix} 12 & 24 \\ 12 & 24 \end{pmatrix}
\quad \Rightarrow \quad d = 0
\quad \Rightarrow \quad t = \pm 6
\quad \Rightarrow \quad A-B = \pm \begin{pmatrix} 2 & 4 \\ 2 & 4 \end{pmatrix} $$
This will give you all the possible 4 choices of the pair $(A, B)$.
Remark. If you want to solve an equation of the form $X^2 = kI$, then the general form of $X$ is
$$ X = \begin{pmatrix}a & b \\ c & -a \end{pmatrix} \quad \text{or} \quad \pm\sqrt{k} I, $$
with $a^2 + bc = k$. So there may be an infinitely many solution.
