Here is an alternative view.
Let $T(x,y)$ be a torsion point on the given curve $E$ over $\Bbb Q$ with (affine) equation
$$ E\ :\qquad Y^2 = X^3 + k\ ,\ k=p^2\ ,\ p>3\text{ prime .}$$
From the OP, we already know that only points $(x,y)$ with $y$ among one of the $12$ divisors in $\Bbb Z$ of $3p^2$ are possible. (Since $y^2$ divides the discriminant $-27k^2=-27p^4$.)
And $y$ determines the value of $x$, since $x\to x^3$ is injective on $\Bbb Z$. The values $\pm p$ and $\pm p^2$ are excluded. So there remain only $8$ possibilities.
The $3$-torsion points are simple $(0,\pm p)$. (Easy computations inserted for the convenience of a community reader only at the end.)
So the torsion order is either $3$ or $6$. In case of a $6$-torsion, we would have a point of $2$-torsion. This is easily excluded, since the polynomial $X^3-p^2$ has no rational roots. So the torsion order is $3$.
$\square$
Addendum: Computation of the $3$-torsion points.
We compute both sides of the equality $T= -2T$.
The point $T$ is $(x,y)$. Let us compute $-2T=(x',y')$.
The slope $m$ in $(x,y)$ comes from the formal differential $2Y\; dY=3X^3\; dX$, it is
$$m=\frac{3x^2}{2y}\ .$$
The components of $-2T$ are thus given by the intersection of the line $(Y-y)=m(X-x)$ with $E$. plugging in $Y = m(X-x)+y$ into the equation of $E$ we have $(m(X-x)+y)^2=X^3+k$, and the three solutions are $x,x,x'$. Vieta for the coefficient in $X^2$ gives $x'$, so
$$
\begin{aligned}
x' &= m^2-x-x=\frac{9x^4}{4y^2}-2x\ ,\\
y' &= y + m(x'-x)\ .
\end{aligned}
$$
From $y'=y$ we get either $x'=x$, so $T=O$, or else $m=0$, so $x=0$. The corresponding points are $T_\pm =(0,\pm A)$, both torsion points. (Opposite to each other.)
Addendum:
An other possibility to eliminate a torsion point of the shape $(x,\pm3)=\pm(x,3)$ is as follows. It is enough to consider only the plus sign, we start with $y=3$, and $P(x,3)$ torsion, and compute $2P=(x',y')$, which is also torsion. We expect an "ugly" expression, but $y'$ is also constrained to be one of the divisors of $3p^2$. With the above notations...
$$
\begin{aligned}
m &=\frac{3x^2}{2y}=\frac{x^2}2\ ,\\
x' &= m^2-2x=\frac {x^4}4-2x\ ,\\
y' &= y+m(x'-x)=3+\frac {x^2}2\left(\frac{x^4}4-3x\right)\\
&=3+\frac{(9-p^2)}2\cdot\frac{(9-p^2)-12}4
=\frac 18(p^4-6p^2-3)
\end{aligned}
$$
The last number is integer and a divisor of $3p^2$. It is $-3/8\ne 0$ in the field $\Bbb F_p$. So it is relatively prime to $p$, thus a divisor of $3$, and there are four cases ($\pm 1$, $\pm 3$) to be checked.
Best Answer
First, notice that an elliptic curve $E/K$ given by \begin{equation}\tag{$\star$} E : y^2 = x^3 + k \end{equation} is isomorphic to the curve $$E' : y^2 = x^3 + u^6 k$$ for any $u \in K^*$ by replacing $x$ by $u^2x$ and $y$ by $u^3y$. Hence by removing $6^{th}$ powers we are looking at
$$E : y^2 = x^3 + q^t$$
where $t = 1, 3, 5$ (in your case - I will deal with all $t = 0, ..., 5$ below).
Now, there is no reason to believe that when $q \equiv 1 \pmod{4}$ that $E/\mathbb{Q}$ will have rank $0$, and in particular when $q = 5$ and $t = 1$ the rank is $\geq 1$ (it is harder to show that equality holds) since $P = (-1, 2)$ has infinite order (this can be seen since e.g., $3P$ is some horrific fraction so Lutz-Nagell would yield a contradiction). Thus I will discard the question on $E(\mathbb{Q})$ and ask it instead for the torsion subgroup $E(\mathbb{Q})_{tors}$.
I'm not sure if one can use Lutz-Nagell to do a case analysis here, but it is certainly harder than using results about $E(\mathbb{Q}_p)$ - so I will do that. Firstly notice that $E(\mathbb{Q}) \subset E(\mathbb{Q}_p)$.
If $q \neq 5$, $E$ has good reduction at $p = 5$ (by your discriminant calculation - although you are missing a factor of $16$). By formal groups $E_1(\mathbb{Q}_5) \cong (\mathbb{Z}_5, +)$ which is torsion free (Silverman AEC IV 6.4 and VII 2.2). Hence $$E(\mathbb{Q})_{tors} \hookrightarrow E(\mathbb{Q}_5)_{tors} \hookrightarrow E(\mathbb{Q}_5)/E_1(\mathbb{Q}_5) \cong \tilde{E}(\mathbb{F}_5)$$
You can check that for curves of the form $(\star)$ have $6$ points. Hence $\#E(\mathbb{Q})_{tors} | 6$.
We deal with $q = 5$ separately. As above in this case $E$ has good reduction at $11$, and we have that $\#E(\mathbb{F}_{11}) = 12$ for each $t = 0, ..., 5$ - so $\#E(\mathbb{Q})_{tors} | 12$.
Thus we only need to check whether $E$ has rational $2$ or $3$ torsion. The 2-torsion is easy, $E$ has 2-torsion if and only if $x^3 + q^t$ has a root. This is the case if and only if $t = 0, 3$, and there is at most $1$ root - hence at most $1$ 2-torsion point.
The $3$-torsion points on $E$ are the inflection points. The Hessian determinant of $X^3 + kZ^3 - Y^2Z$ is $$24X(3q^tZ^2 + Y^2)$$ hence we have affine 3-torsion points if and only if there is a point $(x, y) \in E(\mathbb{Q})$ satisfying $$24x(3q^t + y^2) = 0$$
Since $3q^t > 0$ the second factor cannot be $0$, hence $x = 0$. In that case $y^2 = q^t$, hence $E(\mathbb{Q})$ contains a $3$-torsion point if and only if $t$ is even.
Thus \begin{align*} E(\mathbb{Q})_{tors} &= \begin{cases} \{O\} &t \equiv 1, 5 \pmod{6} \\ \mathbb{Z}/2\mathbb{Z} &t \equiv 3 \pmod{6} \\ \mathbb{Z}/3\mathbb{Z} &t \equiv 2, 4 \pmod{6} \\ \mathbb{Z}/6\mathbb{Z} &t \equiv 0 \pmod{6} \end{cases} \end{align*}
My challenge to you OP, can you generalise the above so that we do not need to assume that $k$ is a prime power?