We can construct a proof based on homotopy invariance, borrowing ideas from the proof of Rouche's theorem.
Let $\gamma : [0, 1] \to\mathbb C$ parametrise the semi-circular contour:
$$ \gamma(t) = \begin{cases} -1 + 4t & t \in [0, \tfrac 1 2] \\ e^{2\pi i (t - \tfrac 1 2 )} & t \in [\tfrac 1 2 , 1]\end{cases}$$
The number of zeroes of $f$ in the upper half disk is equal to the winding number around the origin for the curve $f \circ \gamma : [0,1] \to \mathbb C^\star$:
$$ f \circ \gamma(t) = \begin{cases} (-1 + 4t)^6 + 3(-1+4t)^4+1 & t \in [0, \tfrac 1 2] \\ e^{12\pi i (t - \tfrac 1 2 )} + 3e^{8\pi i (t - \tfrac 1 2 )} + 1 & t \in [\tfrac 1 2 , 1]\end{cases}$$
As you say, this winding number is hard to evaluate. However, since the $3e^{8\pi i (t - \tfrac 1 2 )}$ term is "dominant" for $t \in [\tfrac 1 2 , 1] $, we would expect the winding number of $f \circ \gamma$ around the origin to be the same as the winding number of the simpler-looking curve $g : [0, 1] \to \mathbb C^\star$, which is defined as:
$$ g(t) := \begin{cases} 3 & t\in [0, \tfrac 1 2] \\ 3e^{8\pi i (t - \tfrac 1 2 )} & t \in [\tfrac 1 2 , 1]\end{cases}.$$
To make this intuition rigorous, we exhibit a homotopy $F: [0,1] \times [0,1] \to \mathbb C^\star$ between $g$ and $f \circ \gamma$. A possible homotopy is
$$ F(s , t) = \begin{cases} 3(1-s) + \left((-1 + 4t)^6 + 3(-1+4t)^4+1\right)s & t \in [0, \tfrac 1 2] \\ se^{12\pi i (t - \tfrac 1 2 )} + 3e^{8\pi i (t - \tfrac 1 2 )} + s & t \in [\tfrac 1 2 , 1]\end{cases}$$
The key thing we need to check is that this homotopy avoids the origin, i.e. $F(s, t) \neq 0$ for all $s$ and $t$:
If $t \in [0, \tfrac 1 2 ]$, then $F(s, t) \neq 0$ for all $s \in [0,1]$, because $3$ and $(-1 + 4t)^6 + 3(-1 + 4t)^4 + 1$ are both strictly positive real numbers.
If $t \in [\tfrac 1 2 , 1]$, then $F(s, t) \neq 0$ for all $s \in [0,1]$, because $\left|3e^{8\pi i (t - \tfrac 1 2 )}\right| > \left| e^{12\pi i(t - \tfrac 1 2 )} + 1\right|$.
Thus, having shown that $f \circ \gamma$ and $g$ are homotopic within $\mathbb C^\star$, we deduce that they have the same winding number. As you say, the winding number of $g$ is obviously $2$, so this must be the winding number of $f \circ \gamma$ too.
If $|z|=1$, then $|2z^4|<|5z^2|$ and $|z^4+1|<|10z^2|$. So, both $f$ and $g$ have $2$ zeros when $|z|<1$.
Now, if $|z|=4$, then $|5z^2|<|2z^4|$ and $|10z^2+1|<|z^4|$. So, both $f$ and $g$ have $4$ zeros when $|z|<4$.
Best Answer
You may apply Rouché's theorem directly with respect to the given square $K$.
If $z\in \partial K$ then $|z|^2\geq 3^2=9$ and $$|e^{z-1}|=e^{x-1}\leq e^2$$ where $z=x+iy$ with $-3\leq x\leq 3$ and $-3\leq y\leq 3$. So for any $z\in\partial K$ we have that $$|e^{z-1}|\leq e^2<9\leq |z|^2$$ which implies that $z^2$ and $f(z)=z^2+e^{z-1}$ have the same number of zeros inside $K$, that is $2$.