$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\newcommand\gl{\mathfrak{gl}}\newcommand\sl{\mathfrak{sl}}\newcommand\z{\mathfrak z}\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\tr{tr}$Write $\z_n(k)$ for the space of scalar matrices in $\gl_n(k)$.
There is one more possibility, namely $\mathcal I = \z_2(\mathbb F_2) + \mathbb F_2\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ (which equals $\z_2(\mathbb F_2) + [\SL_2(\mathbb F_2), \SL_2(\mathbb F_2)]$) when $n = 2$ and $k = \mathbb F_2$.
The claim in the question, thus amended, is equivalent to the claim that $\sl_n(k)$ or $\sl_n(k)/\z_n(k)$, as appropriate, is irreducible for $n \ge 2$. Indeed, given the latter claim and a submodule $V$ of $\gl_n(k)$ (still for $n \ge 2$, since $n \le 1$ is obvious), we have one of three possibilities:
$V \cap \sl_n(k)$ is all of $\sl_n(k)$, in which case $V$ contains, hence equals, $\sl_n(k)$; or
$V \cap \sl_n(k)$ is trivial, in which case, for every $X \in V$ and $g \in \SL_n(k)$, we have that $\Ad(g)X - X \in V \cap \sl_n(k)$ equals $0$, so that $V$ is contained in, hence equals, $\z_n(k)$; or
$n = 0$ in $k$ and $V \cap \sl_n(k)$ equals $\z_n(k)$, in which case, for every $X \in V$, we have that $g \mapsto \Ad(g)X - X$ may be viewed as a homomorphism $\SL_n(k) \to k$. If the homomorphism is trivial, which is automatic when $n > 2$ or $k \ne \mathbb F_2$ (since then $\SL_n(k)$ is its own derived group), then again $V$ equals $\z_n(k)$. If $n = 2$, $k = \mathbb F_2$, and $V$ is not contained in $\sl_n(k)$, then direct computation shows that the only coset of $\z_2(k)$ affording the unique non-trivial homomorphism $\SL_2(k) \to k$ is $\z_2(k) + \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$. In particular, $V$ is contained in $\mathcal I$, hence equals $\mathcal I$ or $\z_2(k)$.
I originally intended to mimic the highest-weight argument of Proposition 1.10 of Liebeck and Seitz - On the subgroup structure of exceptional groups, but I realised that weights for diagonal matrices in $\SL_n(\mathbb F_2)$ don't tell us much. Fortunately, just the "highest" part is enough.
Write $U$ for the subgroup of $\SL_n(k)$ consisting of matrices that are "upper unitriangular" (upper triangular, with $1$s on the diagonal).
Let $W$ be a non-$0$ submodule of $\sl_n(k)/(\z_n(k) \cap \sl_n(k))$. By Engel's theorem, the space of $U$-fixed vectors in $W$ is non-$0$. By direct computation, the $U$-fixed subspace of $\sl_n(k)/(\sl_n(k) \cap \z_n(k))$ is spanned by the matrix with $1$ in the $(1, n)$ position, and $0$ elsewhere. To ease notation, let's write $E_{1, n}$ for this matrix, with analogous notation $E_{i, j}$ in general. Then we have shown that the image of $E_{1, n}$ belongs to $W$.
Conjugation by appropriate signed permutation matrices (followed by multiplication by $\pm$) shows that the image of each $E_{i, j}$ with $i \ne j$ also belongs to $W$. Similarly, the image of $\Ad(1 + E_{n, 1})E_{1, n} = E_{1, 1} - E_{n, n}$ belongs to $W$, so again conjugation by signed permutation matrices shows that the images of all $E_{i, i} - E_{j, j}$ belong to $W$. Since $\{E_{i, i} - E_{j, j}, E_{i, j} \mathrel: 1 \le i \ne j \le n\}$ spans $\sl_n(k)/(\sl_n(k) \cap \z_n(k))$, we are done.
