Let $k=r(B_n,B_n)-1$. Then we can color the edges of $K_k$ red and blue in such a way that there is no monochromatic $B_n$. Let $E_r$ and $E_b$ be the sets of red and blue edges, respectively, let $V\,$ be the vertex set of $K_k$, and for $v\in V\,$ let $\deg_r(v)$ and $\deg_b(v)$ be the number of red and blue edges, respectively, incident at $v$.
For each edge $e=\{u,v\}$ let $M(e)$ be the set of vertices that are connected to both $u$ and $v$ by edges of the same color as $e$, and let $\mu(e)=|M(e)|$. Then the number of monochromatic triangles in $K_k$ is $$m=\frac13\sum_{e\in E_r\cup E_n}\mu(e)\;,$$ since each triangle gets counted three times.
Suppose that $e=\{u,v\}\in E_r$. The edge $e$ together with the edges $\{u,w\}$ and $\{v,w\}$ for $w\in M(e)$ yield a red $B_{\mu(e)}$, so we must have $\mu(e)
Let $v\in V$. Each pair $\langle e_r,e_b\rangle\in E_r\times E_b$ such that $e_r$ and $e_b$ are both incident at $v$ determines a non-monochromatic triangle in $K_k$, so $v$ is a vertex of $\deg_r(v)\deg_b(v)$ non-monochromatic triangles. Each non-monochromatic triangle has two such vertices, so the number of non-monochromatic triangles is $K_k$ is $$\frac12\sum_{v\in V}\deg_r(v)\deg_b(v)\;,$$ and
$$\begin{align*}m&=\binom{k}3-\frac12\sum_{v\in V}\deg_r(v)\deg_b(v)\\
&=\binom{k}3-\frac12\sum_{v\in V}\deg_r(v)(k-1-\deg_r(v))\\
&=\binom{k}3-\frac{k-1}2\sum_{v\in V}\deg_r(v)+\frac12\sum_{v\in V}\deg_r(v)^2\\
&=\binom{k}3-(k-1)|E_r|+\frac12\sum_{v\in V}\deg_r(v)^2\;.
\end{align*}$$
By Cauchy’s inequality we have $$\left(\sum_{v\in V}\deg_r(v)\right)^2\le\sum_{v\in V}\deg_r(v)^2\; \sum_{v\in V}1^2=k\sum_{v\in V}\deg_r(v)^2\;,$$ so $$4|E_r|^2\ge k\sum_{v\in V}\deg_r(v)^2\;,$$ and
$$\begin{align*}m&\ge\binom{k}3-(k-1)|E_r|+\frac{2|E_r|^2}k\\
&=\binom{k}3-|E_r|\left(k-1-\frac{2|E_r|}k\right)\\
&=\binom{k}3-|E_r|\frac2k\left(\frac{k(k-1)}2-|E_r|\right)\\
&=\binom{k}3-|E_r|\frac2k\left(\binom{k}2-|E_r|\right)\\
&=\binom{k}3-|E_r|\frac2k|E_b|\;.
\end{align*}$$
Letting $d=\frac12\big||E_r|-|E_b|\big|$, we then have
$$\begin{align*}
m&\ge\binom{k}3-\frac2k\left(\frac12\binom{k}2-d\right)\left(\frac12\binom{k}2+d\right)\\
&=\frac{k(k-1)(k-2)}6-\frac2k\left(\frac14\binom{k}2^2-d^2\right)\\
&=\frac{k(k-1)(k-2)}6-\frac{k(k-1)^2}8+\frac{2d^2}k\\
&=\frac{k(k-1)(k-5)}{24}+\frac{2d^2}k\;,
\end{align*}$$
and combining this with $(1)$ yields
$$\begin{align*}
\frac{k(k-1)(k-5)}{24}+\frac{2d^2}k&\le\frac13(n-1)\binom{k}2\\
&=\frac{k(k-1)(n-1)}6\;,
\end{align*}$$
or
$$\begin{align*}
\frac{2d^2}k&\le\frac{k(k-1)(n-1)}6-\frac{k(k-1)(k-5)}{24}\\
&=\frac{k(k-1)(4n+1-k)}{24}\;.
\end{align*}$$
Then $4n+1-k\ge 0$, so $r(B_n,B_n)=k+1\le 4n+2$.
This is actually a special case of Theorem 1.5 of Li & Zang, Introduction to Graph Ramsey Theory.