= Solution
In fact, \b[$2,3,5,7$ must all be primes of bad reduction].
At a prime of <good reduction of an elliptic curve>, the kernel $E_1(\mathbb Q_p)$ of reduction is the <formal group of an elliptic curve> evaluated on $p\mathbb Z_p$, using its parameter at the identity. This kernel has no prime-to-$p$ torsion: for an integer $\ell$ prime to $p$, the multiplication series is $[\ell]_F(T)=\ell T+$ higher-degree integral terms, so for nonzero $T\in p\mathbb Z_p$ its first term has strictly smaller valuation than the others and $[\ell]_F(T)\ne0$.
For each odd good prime, the entire given torsion subgroup, of order $16$, would therefore inject into $E(\mathbb F_p)$. But the <Hasse theorem for elliptic curves> gives
$$
\#E(\mathbb F_3)\leq7,\qquad\#E(\mathbb F_5)\leq10,\qquad\#E(\mathbb F_7)\leq13.
$$
None can contain a subgroup of order $16$. Hence $3,5,7$ are bad.
For the <prime-two torsion bound for good reduction>, use the deeper formal subgroup corresponding to $4\mathbb Z_2$. Part (i), with $m=2>1/(2-1)$, shows that $E_2(\mathbb Q_2)$ is torsion-free and has index $2$ in $E_1(\mathbb Q_2)$. Therefore any finite torsion subgroup of $E_1(\mathbb Q_2)$ injects into $E_1/E_2$ and has order at most $2$.
If $E$ had good reduction at $2$, its rational torsion subgroup of order $16$ would consequently have reduction image of order at least $8$. The <Hasse theorem for elliptic curves> instead gives $\#E(\mathbb F_2)\leq5$. This contradiction proves bad reduction at $2$ as well, and therefore
$$
\boxed{\#\{\text{bad reduction primes of }E\}\geq4.}
$$
The special formal-group argument at $2$ is necessary; prime-to-$p$ injectivity alone says nothing about the given $2$-primary torsion there.
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