WLOG, assume that $n\ge m$. The answer is $k=\min{(2n+2m+2,3n+2)}$. For $k\ge 2n+2m+3$, a path of length $2n+2m+3$ is a counterexample. For $k\ge 3n+3$, take 3 disjoint paths with $n+1,n+1,n$ vertices, and connect these paths with a new vertex from their endpoints. Now assume $k\le \min{(2n+2m+2,3n+2)}$ and we can't choose two such vertices. Let $A_1A_2\dots A_t$ be a diameter of the tree and let $G_i$ be the subtree of $A_i$. (not intersecting with the diameter) If $t\le 2n$, then choosing $v=A_{\lfloor \frac{t}{2} \rfloor}$ works. Assume $t>2n$. Choosing $v=A_{n+1}$ and $u=A_{t-m}$, there exists a vertex $a$ such that $d(a,v)>n$ and $d(a,u)>m$. If $a\in G_i$, then $n+1<i<t-m$ by the diameter assumption. Choosing $v=A_{t-n}$ and $u=A_{m+1}$, there exists a vertex $b$ such that $d(b,v)>n$ and $d(b,u)>m$. If $b\in G_j$, then $m+1<j<t-n$ by the diameter assumption. If $i=j$, then we have $n+1<i<t-n$. This means there are at least $n+n+d(A_{n+1},a)+2 \ge 3n+3$ vertices, contradiction. If $i\neq j$, then let $x=d(a,A_i)$ and $y=d(b,A_j)$. Summing up the inequalities $x+d(A_i,A_{n+1})\ge n+1$ and $x+d(A_{t-m},A_i)\ge m+1$, we get $2x+(t-m-n-1)\ge m+n+2$. Similarly we get $2y+(t-m-n-1)\ge m+n+2$. Finally, summing the last two we get $x+y+t\ge 2n+2m+3$. But since $i\neq j$, there are at least $x+y+t$ vertices in the tree, contradiction.