Let $|G|=2^n \cdot 3, \ \ n \geqslant 2$ and $H$ Sylow 2-subgroup of $G$. Then $|G:H|=3$, so $|G:\operatorname{Core}(H)| | 3!=6$. So, $\operatorname{Core}(H)$ is non-trivial group, because $4| |G|$ and $4 \nmid6.$ $\operatorname{Core}(H)$ is normal, and she is not whole group because $\operatorname{Core}(H) \le H \lneq G$. So, $G$ is not simple group.
A simpler approach, let $P \in Syl_3(G)$ and assume $G$ to be simple. Then $|G:P|=4$. $G$ acts by left multiplication on the left cosets of $P$ and the kernel of this action is core$_G(P)=\bigcap_{g \in G}P^g$. This is a normal subgroup of $G$ and since $G$ is simple, it it trivial. This means that $G$ embeds isomorphically into $S_4$. This implies $36 \mid 24$, by Lagrange's Theorem, a contradiction. Hence $G$ cannot be simple.
Note on the solution The following is a generalization of the well-known Cayley imbedding, which can be recovered by taking $H=1$ in the sequel.
Let $H \leq G$, with $|G:H|=n$ a positive integer. Put $\Omega=\{g_1H,g_2H, \ldots ,g_nH\}$, the set of left cosets of $H$. $G$ works on $\Omega$ by multiplication from the left: if $g \in G$, then $g(g_iH)=gg_iH$ is again one of the elements of $\Omega$. So each $g \in G$ induces a permutation $\sigma_g$ on $\Omega$. Hence we can define a map $f: G \rightarrow S_n$, by $f(g)=\sigma_g$. It is easy to verify that $f$ is a homomorphism (observe that $\sigma_g \circ \sigma_h=\sigma_{gh}$ for all $g,h \in G$). Now let us compute ker$(f)$. If $x \in $ ker$(f)$, this means that $x$ induces the trivial permutation, so for any left coset $gH$, we must have $xgH=gH$, implying $x \in gHg^{-1}$ for all $g \in G$. This amounts to $x \in \underset{g \in G}\bigcap H^g$ (where $H^g=g^{-1}Hg$; note that when $g$ runs through the entire $G$, $g^{-1}$ does the same, so intersection over $g^{-1}Hg$ is the same as over $gHg^{-1}$). Conversely, if $x \in \underset{g \in G}\bigcap H^g$, then it induces a trivial permutation.
We conclude that ker$(f)=\underset{g \in G}\bigcap H^g$ and by the first isomorphism theorem $G/$ker$(f) \cong $ im$(f)$, being a subgroup of $S_n$.
In general, if $H \leq G$, the core$_G(H)$ is the largest normal subgroup of $G$ contained in $H$. It is easy to show that core$_G(H)=\underset{g \in G}\bigcap H^g$.
Best Answer
I would avoid the counting argument, since there are indeed groups of order $60$ with $15$ Sylow $2$-subgroups, and all elements of order $2$ are shared by several of these Sylow subgroups at once.
Here is a slightly easier way to do this, using a slightly stronger version of Sylow's theorems. Namely,
Theorem: Let $p$ be a prime, and $p^r$ the highest power of $p$ dividing $G$. For any two Sylow $p$-subgroups $P$ and $Q$ of $G$, suppose we have $p^k\le |P\cap Q|$. Then the number of Sylow $p$-subgroups of $G$ - call it $n_p$ - satisfies $n_p\equiv 1\pmod{p^{r-k}}$.
The proof is exactly the same as the usual proof (using group actions), you just have to pay extra attention to orbit sizes along the way.
Now, if $G$ is a group of order $60$, containing $15$ Sylow $2$-subgroups, then since $15\not\equiv 1\pmod{4}$, the theorem implies the existence of two Sylow $2$-subgroups - call them $P$ and $Q$ - with $|P\cap Q|=2$. But then $P\cap Q$ is normal in both $P$ and $Q$, and thus its normalizer, $N_G(P\cap Q)$, has order divisible by $4$, and size at least $|P|+|Q|-|P\cap Q|=6$. Thus $N_G(P\cap Q)$ has size at least $12$.
The action of $G$ on the right cosets of $N_G(P\cap Q)$ gives a homomorphism from $G$ into $S_5$. If we are assuming $G$ is simple, this becomes an isomorphism with $A_5$.
[Note that this is actually a contradiction, since $A_5$ does not have $15$ Sylow $2$-subgroups, but either way, you are done.]