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). Similarly, $\mu(e) for each $e\in E_b$. Thus, $m\le\frac13(n-1)|E_r\cup E_b|=\frac13(n-1)\binom{k}2\;.\tag{1}$
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.