2. Given $f(x)=x^{2}+2 x+1$, there exists a real number $t$ such that when $x \in[1, m]$, $f(x+t) \leqslant x$ always holds, then the maximum value of $m$ is $\qquad$ .
2.4.

Translate the graph of $f(x)$ to the right by $-t$ units, and from the graphical analysis, the maximum value of $m$ is the larger of the x-coordinates of the two intersection points of $\left\{\begin{array}{l}y=x, \\ y=f(x+t)\end{array}\right.$
From $f(1+t)=1$, we get $t=-1, t=-3$.
Then, from $f(x-3)=x$, we have
$x=1$ (discard), $x=4$.
Therefore, the maximum value of $m$ is 4.