7. Let $\left\{a_{n}\right\}$ be a monotonically increasing sequence of positive integers, satisfying
$$
a_{n+2}=3 a_{n+1}-a_{n}, a_{6}=280 \text {. }
$$

Then $a_{7}=$
7.733.

From the problem, we have
$$
\begin{array}{l}
a_{3}=3 a_{2}-a_{1}, \\
a_{4}=3 a_{3}-a_{2}=8 a_{2}-3 a_{1}, \\
a_{5}=3 a_{4}-a_{3}=21 a_{2}-8 a_{1}, \\
a_{6}=3 a_{5}-a_{4}=55 a_{2}-21 a_{1}=280=5 \times 7 \times 8, \\
a_{7}=3 a_{6}-a_{5}=144 a_{2}-55 a_{1} .
\end{array}
$$

Since $(21,55)=1$, and $a_{2}>a_{1}$ are both positive integers, it follows that $5\left|a_{1}, 7\right| a_{2}$.
Let $a_{1}=5 b, a_{2}=7 c\left(b, c \in \mathbf{Z}_{+}\right)$, we get $11 c-3 b=8$.
It is easy to verify that $a_{1}=5, a_{2}=7$.
Then $a_{7}=144 a_{2}-55 a_{1}=733$.