Chọn phương án D.

Đặt $\begin{cases}
\cos\alpha=\dfrac{3}{5}\\
\sin\alpha=\dfrac{4}{5}
\end{cases}$. Ta có $$\begin{aligned}
y&=3\sin x+4\cos x+1\\
\Leftrightarrow\dfrac{y}{5}&=\dfrac{3}{5}\sin x+\dfrac{4}{5}\cos x+\dfrac{1}{5}\\
&=\cos\alpha\sin x+\sin\alpha\cos x+\dfrac{1}{5}\\
&=\sin\left(x+\alpha\right)+\dfrac{1}{5}\\
\Leftrightarrow y&=5\sin\left(x+\alpha\right)+1.
\end{aligned}$$
Mặt khác,
\begin{eqnarray*}
-1\leq&\sin\left(x+\alpha\right)&\leq1\\
\Leftrightarrow-5\leq&5\sin\left(x+\alpha\right)&\leq5\\
\Leftrightarrow-4\leq&5\sin\left(x+\alpha\right)+1&\leq6\\
\Leftrightarrow-4\leq&y&\leq6.
\end{eqnarray*}
Vậy $\begin{cases}
M=6\\ m=-4.
\end{cases}$