Is $x^5 + x^3 + 1$ irreducible in $\mathbb{F}_{32}$ and $\mathbb{F}_8$?

Let $g(x)$ be an irreducible factor of $f(x)$ over $\mathbb F_8$, so $f(x)=g(x)h(x)$ for some $h(x)\in\mathbb F_8[x]$. From the inclusion of fields $\mathbb F_2\rightarrow\mathbb F_8$ we obtain an inclusion $\mathbb F_2[x]\rightarrow\mathbb F_8[x]$, giving a homomorphism $$ \phi:\mathbb F_2[x]\rightarrow\mathbb F_8[x]/g(x). $$ Now $\phi(f(x))=g(x)h(x)+g(x)\mathbb F_8[x]=0+g(x)\mathbb F_8[x]$, so $\phi$ induces a homomorphism $$ \bar\phi:\mathbb F_2[x]/f(x)\rightarrow\mathbb F_8[x]/g(x). $$ Since $f(x)$ is irreducible over $\mathbb F_2$ and $g(x)$ over $\mathbb F_8$, we have $$ \mathbb F_2[x]/f(x)\cong\mathbb F_{32},\hspace{10mm} \mathbb F_8[x]/g(x)\cong\mathbb F_{8^d} $$ where $d=\deg g(x)$. Therefore $\bar\phi$ gives a homomorphism $$ \mathbb F_{32}\rightarrow\mathbb F_{8^d}. $$ In particular $\mathbb F_{8^d}$ is a vector space over $\mathbb F_{32}$, so $$ 5=\log_2(32)\mid\log_2(8^d)=3d. $$ Hence $5|d$, so $d\geq5$. Thus $f(x)$ must be a scalar multiple of $g(x)$, and in particular $f(x)$ is irreducible over $\mathbb F_8$.

Here’s another approach, based on your good start. You’ve observed that $f$ is $\Bbb F_2$-irreducible, and that indeed $\Bbb F_{32}$ is its splitting field. In other words, $f(x)=\prod_i(x-\alpha_i)$, where the alpha’s are five different elements of $\Bbb F_{32}$. If you could take fewer than five of the factors above and multiply them together to get a polynomial in $\Bbb F_8[x]$, then its coefficients would be in $\Bbb F_{32}\cap\Bbb F_8=\Bbb F_2$, in other words a proper $\Bbb F_2$-divisor of $f$. Thus the only way you can multiply some of those factors together to get an $\Bbb F_8$-polynomial is to use all of them. So $f$ is $\Bbb F_8$-irreducible.