Prove $m=3k+1 \quad m,k \in \mathbb Z \implies m^2=3l+1 \quad m,l \in \mathbb Z$

Suppose we call an integer “throdd” $\iff$ $m=3k+1$ for some integer
$k$. Prove that the square of any throdd integer is throdd.

So here is what I have so far:

$$(3k+1)^2 = 3k+1$$
$$(3k+1)(3k+1) = 3k+1$$

Am I going in the right direction?
2nd part: $(3k+1)(3k+1)= 9k^2 + 6k + 1$

$3k(3k+ 2) +1$
So im confused because $(3k +2)$ is not m and $3k(3k+2) +1$ isnt in the form of $3m+1$?

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