Suppose that $A_{1},A_{2},\ldots,A_{m}$ are distinct $n$-by-$n$ real matrices such that $A_{i}A_{j}=0$ for all $i\neq j$. Then, prove that $m\leq n$.
It is clear that matrices are not invertible. Then columns of any matrix are linearly dependent. From this we can construct other matrices.
On
Call the images and kernels of the matrices $I_i$ and $K_i$ and assume $m>n$.
We prove $I_1+\dots+I_n=\mathbb R^n$ by showing a linearly independent set $a_1,\dots,a_n$ exists with $a_i\in I_i$.
We build it inductively.
Suppose that $a_1,\dots,a_i$ span $I_i$, since all of the $a_i$ are in $K_i$ we have $K_i\geq I_i$, a contradiction.
Hence there is a suitable $a_i$ and we are done.
On the other hand, we have $I_1+\dots+I_n\subseteq K_{n+1}$ which is a contradiction.
Remark: In this answer, there is no extra condition on $\mathbf{A}_i^2$, where $i=1,2,\ldots,m$, just as the OP initially stated. They can be zero.
I shall work with the assumption that the ground field can be any field $K$ which is not an imperfect field of characteristic $2$. If you assume that $\mathbf{A}_1$, $\mathbf{A}_2$, $\ldots$, $\mathbf{A}_m$ are linearly independent instead of just being distinct, then $$m\leq \max\left\{n,\left\lfloor\frac{n^2}{4}\right\rfloor\right\}\tag{*}$$ and this bound is sharp for every $n\in\mathbb{Z}_{>0}$.
Clearly, the subalgebra $S$ of the matrix algebra $M:=\text{Mat}_{n\times n}(K)$ generated by $\mathbf{A}_1,\mathbf{A}_2,\ldots,\mathbf{A}_m$ is a commutative subalgebra of $M$. For two subalgebras $U$ and $V$ of $M$, we say that $U$ is conjugate to $V$ if there exists an invertible element $\mathbf{E}$ of $M$ such that $V=\mathbf{E} U\mathbf{E}^{-1}$, namely, $$V=\big\{\mathbf{E}\mathbf{P}\mathbf{E}^{-1}\,\big|\,\mathbf{P}\in U\big\}\,.$$
By Theorem 2 of this paper, $m\leq \left\lfloor\frac{n^2}{4}\right\rfloor+1$. For $n>3$, according to Theorem 3 of the same paper, the equality $m=\left\lfloor\frac{n^2}{4}\right\rfloor+1$ holds if and only if $S$ is conjugate to the subalgebra $B$ containing all matrices of the form $$\alpha \,\mathbf{I}+\begin{bmatrix}\boldsymbol{0}_{\left\lfloor\frac{n}{2}\right\rfloor\times\left\lfloor\frac{n}{2}\right\rfloor}&\mathbf{X}\\\boldsymbol{0}_{\left\lceil\frac{n}{2}\right\rceil\times \left\lfloor\frac{n}{2}\right\rfloor}&\boldsymbol{0}_{\left\lceil\frac{n}{2}\right\rceil\times \left\lceil\frac{n}{2}\right\rceil}\end{bmatrix}\,,$$ where $\alpha \in K$, $\textbf{I}$ is the $n$-by-$n$ identity matrix, $\boldsymbol{0}_{r\times s}$ is the $r$-by-$s$ zero matrix, and $\mathbf{X}$ is an $\left\lfloor \frac{n}{2}\right\rfloor$-by-$\left\lceil\frac{n}{2}\right\rceil$ matrix, or $S$ is conjugate to the subalgebra $\bar{B}$ consisting of all matrices of the form $$\bar{\alpha} \,\mathbf{I}+\begin{bmatrix}\boldsymbol{0}_{\left\lceil\frac{n}{2}\right\rceil\times \left\lceil\frac{n}{2}\right\rceil}&\bar{\mathbf{X}}\\\boldsymbol{0}_{\left\lfloor\frac{n}{2}\right\rfloor\times \left\lceil\frac{n}{2}\right\rceil}&\boldsymbol{0}_{\left\lfloor\frac{n}{2}\right\rfloor\times\left\lfloor\frac{n}{2}\right\rfloor}\end{bmatrix}\,,$$ where $\bar{\alpha}\in K$ and $\bar{\mathbf{X}}$ is an $\left\lfloor\frac{n}{2}\right\rfloor$-by-$\left\lceil\frac{n}{2}\right\rceil$ matrix. However, clearly, there are no basis $$\left\{\mathbf{Y}_0,\mathbf{Y}_1,\mathbf{Y}_2,\ldots,\mathbf{Y}_{\left\lfloor\frac{n^2}{4}\right\rfloor}\right\}$$ of $B$ or of $\bar{B}$ with the property that $\mathbf{Y}_i\mathbf{Y}_j=\textbf{0}_{n\times n}$ for all $i,j\in\left\{0,1,2,\ldots,\left\lfloor\frac{n^2}{4}\right\rfloor\right\}$ with $i\neq j$, since one of the $\mathbf{Y}_i$ must take the form $y\mathbf{I}+\mathbf{N}$ with $y\neq 0$ and $\mathbf{N}\in M$ is strictly upper triangular. Thus, for $n>3$, $$m \leq \left\lfloor\frac{n^2}{4}\right\rfloor\,.$$
For $n\in\{1,2,3\}$, we have $$n=\left\lfloor\frac{n^2}{4}\right\rfloor+1\,,$$ so the inequality (*) is true. Furthermore, for $n\in\{1,2,3\}$, we can take $S$ to be the subalgebra of $M$ consisting of diagonal matrices and find a suitable basis. For $n>3$, we can take $S$ to be the subalgebra of $B$ consisting of strictly triangular matrices and yield $m=\left\lfloor\frac{n^2}{4}\right\rfloor$, and any basis of this subalgebra will work.
P.S.: If Jorge Fernández is right, then the OP forgot to add the condition that $\mathbf{A}_i^2$ is nonzero for all $i=1,2,\ldots,m$. With this condition, $m\leq n$ holds, and a proof can be found here.