A topological space $X$ is called reducible if $X=X_1 \cup X_2$ where $X_1, X_2$ are non-empty, proper subsets of $X$ and closed.
An irreducible algebraic set in $\mathbb{A}^n$ is called an affine variety.
$\varnothing = Z(1)$, so $\varnothing$ is an algebraic set, and obviously $\varnothing \subseteq \mathbb{A}^n$.
If we suppose that $\varnothing$ is reducible, then the only options for $X_1$ and $X_2$ are $\varnothing$, but this can't happen, because $X_1$ and $X_2$ must be non-empty.
So, am I right saying that
$\varnothing$ is an affine variety?
With your definition of "irreducible" and "affine variety", the answer is yes.
However, there are good reasons for considering this to be the wrong definition of "irreducible", as explained in this question. Here's another: the defining property of sober spaces (examples include algebraic varieties with the Zariski topology and Hausdorff spaces) is "every irreducible closed set has a unique generic point". This is nonsense if we allow $\emptyset$ to be irreducible.