No. Suppose on the contrary that $f(x)$ is reducible. Then so is
$$4^{2n} f(x) = (4x^2 + 4x)^{2n} + 4^{2n} = ((2x + 1)^2 - 1)^{2n} + 4^{2n}.$$
Let $y = 2x + 1$. Then we have $x = \frac{y-1}{2}$, and hence
$$g(y) = (y^2 - 1)^{2n} + 4^{2n} \quad (1)$$
is reducible over $\mathbb{Q}$. By Gauss's lemma, $g$ is reducible over $\mathbb{Z}$.
We claim that $g$ can be factorized into even polynomials over $\mathbb{Z}$. Consider a non-constant irreducible factor $h$ of $g$. We are done if $h$ is an even polynomial (since then $\frac{g}{h}$ must be even). Assume $h$ is not even. As $h(y) \mid g(y)$, we have $h(-y) \mid g(-y) = g(y)$. Since $h$ is irreducible and is not even, $h(y)$ and $h(-y)$ must be relatively prime. It follows that
$$g(y) = h(y)h(-y)p(y)$$
for some $p \in \mathbb{Z}[y]$. Note that $h(y)h(-y)$ is an even polynomial. So we are done unless $p$ is a constant polynomial. Let $p(y) = c$, and let $a$ be the leading coefficient of $h$. Then the leading coefficient of $g$ is $a^2c$. In view of (1), we must have $c = 1$ and $a = \pm 1$. Let $b$ be the constant term of $h$. Then the constant term of $g$ is $b^2c = b^2$. Comparing with (1), we obtain $4^{2n} + 1 = b^2$. This implies $4^{2n} = (4^{2n-1})^2$ and $b^2$ are consecutive perfect squares, which is impossible as $4^{2n} > 1$. Therefore, we have proven that $g$ can be factorized into even polynomials over $\mathbb{Z}$.

As a result, by letting $z = y^2$, we find that $(z - 1)^{2n} + 4^{2n}$ is also reducible. Let $z = 4t + 1$. This shows $4^{2n}(t^{2n} + 1)$ is reducible, and so are $t^{2n} + 1$ and $(s+1)^{2n} + 1$. However, note that
$$(s+1)^{2n} + 1 = s^{2n} + \sum_{r=1}^{2n-1} \binom{2n}{r} s^r + 2.$$
The leading coefficient is odd, while all other coefficients are even. Also, the constant term is not a multiple of 4. By Eisenstein's criterion, we know that $(s+1)^{2n} + 1$ is irreducible. This is a contradiction. Therefore, $f(x)$ is irreducible.