How to find the remainder of dividing polynomial $x^{2016}-x^{2015}-1$ with $x^2+1$

Your approach works except that it is not fully general: because $x^2+1$ is quadratic, the quotient-remainder form should be $$ P(x) = Q(x)(x^2+1) + ax+b $$ for some unknown $a$ and $b$.

Now we can set $P(i) = Q(i)(0) + ai + b$, getting one equation for $a$ and $b$. We can also set $P(-i) = Q(i)(0) + a(-i) + b$, getting another equation, because $x^2+1$ is $0$ when $x=i$ or when $x=-i$.

Solving for $a$ and $b$, you should get $a=1$ and $b=0$.

The remainder of $P(x)$ with $Q(x)$ stays invariant if we add or subtract a multiple (which can be a polynomial) of $Q(x)$ from $P(x)$. So, we can evaluate the remainder in the analogous way as we would do in case if ordinary numbers using modular arithmetic.

We thus need to do the computations modulo $Q(x) = x^2 + 1$. Since:

$$x^2 = -1 \bmod Q(x)$$

we have:

$$x^{2016} = 1 \bmod Q(x)$$

$$x^{2015} = x^3 \bmod Q(x) = -x \bmod Q(x)$$

Therefore:

$$x^{2016} - x^{2015} - 1 = x \bmod Q(x)$$

$\!\bmod\, x^{\large 2}\!+\!1\!:\,\ x^{\large 2}\equiv -1\,\Rightarrow\, \color{#0a0}{x^{\large 3}\equiv -x}\,\Rightarrow\,\color{#c00}{x^{\large 4}\equiv 1}\,\Rightarrow\, x^{\large\color{#c00}4q+r} = (\color{#c00}{x^{\large 4}})^{\large q} x^{\large r} \equiv\, x^{\large r}$

$\begin{align}{\rm Therefore}\ \ \ \qquad x^{\large\color{#c00} 4n}\! - x^{\large \color{#c00}4k\color{#0a0}{+3}} &-\, 1\\ \equiv\quad\ 1\ \ \,-\,\ \ \color{#0a0}{\large x^3} &-\, 1\, \equiv\, \color{#0a0}x\ \end{align}$

Remark $ $ Above $x$ like $i$ is a square root of $-1$. If you know a little ring theory you know how to say this much more precisely $\,\Bbb Z[x]/(x^2+1)\cong \Bbb Z[i],\,$ so replacing $x$ by $i$ above gives an isomorphic calculation in the Gaussian integers. So we can use numbers to deduce results about polynomials!