$2\mid n^{n-1}-1$, so $n$ must be odd, so by Lifting The Exponent Lemma (LTE) (this particular lemma is easy to prove) $\upsilon_2\left(n^{n-1}-1\right)=\upsilon_2(n-1)+\upsilon_2(n+1)+\upsilon_2(n-1)-1=2015$. Either $4\mid n-1$ or $4\mid n+1$. $1)$ If $4\mid n-1$, then $\upsilon_2(n+1)=1$, so $2\upsilon_2(n-1)=2015$, impossible (LHS is even and RHS is odd). $2)$ If $4\mid n+1$, then $\upsilon_2(n-1)=1$, so $\upsilon_2(n+1)=2014$. Answer: $n+1=2^{2014}m$ for odd $m\in\Bbb Z^+$.

This is straightforward if one knows Lifting the exponent lemma. Even without lifting the exponent lemma, one can do it easily. Assume $n-1=2^rl$ with $l$ odd. Let $N=n^l$. \[n^{n-1}-1=N^{2^r}-1=\left(N^{2^{r-1}}+1\right)\cdots(N+1)(N-1)\]Clearly, $n$ is odd, so is $N$. For $r>1$, $N^{2^{r-1}}+1\equiv2\pmod4$, so each of them contributes exactly one $2$. $N-1=(n-1)(n^{l-1}+\ldots+1)=S(n-1)$ and since $l$ is odd, $S$ is odd too. Therefore, $N-1$ contributes the same number of $2$ as $n-1$. Same goes for $N+1=n^l+1$. If $r>1$, then $N+1$ gives exactly one $2$. Then the maximum power of $2$ in the expression is $2r$, which is even. This forces that $r=1$. Now, it's just calculation.

So by the Lifting The Exponent Lemma, we have $$2015=v_2(n^{n-1}-1)=2v_2(n-1)+v_2(n+1)-1.$$As $\gcd(n+1,n-1)=2$, we have either $v_2(n-1)=1$ or $v_2(n+1)=1$. If $v_2(n+1)=1$, then $2v_2(n-1)=2015$, which is absurd by $\pmod 2$. If $v_2(n-1)=1$, then $v_2(n+1)=2014$, thus $\boxed{n=2^{2014}k-1}$ for any odd $k$.

When the problem screams LTE so hard. Clearly $2 \mid n-1$ so by LTE $$2015=v_2(n^{n-1}-1)=2v_2(n-1)+v_2(n+1)-1 \implies 4 \mid n+1 \implies v_2(n+1)=2014$$Hence any number of the form $n=2^{2014} \cdot m-1$ where $m$ is odd, works thus we are done