$\def\Spin{\mathrm{Spin}}$The spin group has a representation $S$ called the spin representation, which has dimension $2^m$ for $N=2m$ or $2m+1$. It is irreducible for $N$ odd and splits into two complex conjugate representations $S_+ \oplus S_-$ for $N$ even, with each of $S_{\pm}$ being of dimension $2^{m-1}$. These are the smallest representations of $\Spin(N)$ which do not factor through $SO(N)$. So, once the power of $2$ exceeds $N$, the spin group will not embed in $SU(N)$.

This leaves us to analyze $1 \leq N \leq 6$, and $N=8$. Each of these has some special form.

$\Spin(1)$ is the trivial group. Clearly, this embeds in $SU(1)$.

$\Spin(2)$ is the circle group $\mathrm{R}/\mathrm{Z}$. Note that $SO(2)$ and $\mathrm{Spin}(2)$ are isomorphic as abstract groups, but the map $\mathrm{Spin}(2) \to SO(2)$ is the double cover. Clearly, this embeds in $SU(2)$.

When $N=3$, the spin representation $S$ is two dimensional and gives an isomorphism $\Spin(3) \cong SU(2)$. So, in particular, $\Spin(3)$ embeds in $SU(3)$.

When $N=4$, the representations $S_+$ and $S_-$ are two dimensional, giving an isomorphism $\Spin(4) \cong SU(2) \times SU(2)$. Conveniently, this embeds into $SU(4)$. We cannot use just one of the two $SU(2)$ representations, as the individual projections are not convenient. Particle physicists call working with $\Spin(4)$ in the form $SU(2) \times SU(2)$ the "spinor helicity formalism". If you talk to the sort of particle physicists who care what the signature of spacetime is, you'll want to know that $\Spin(3,1) \cong SL_2(\mathbb{C})$.

When $N=5$, $\dim S = 4$. Indeed, this $4$-dimensional vector space can be identified with a $2$-dimensional quaternionic vector space and $\Spin(5)$ is the group of $2 \times 2$ unitary quaternion matrices. So $\Spin(5)$ embeds into $SU(4)$ (and also into $SU(5)$.)

When $N=6$, we have $\dim S_+ = 4$ and gives an isomorphism $\Spin(6) \cong SU(4)$. So $\Spin(6)$ embeds into $SU(6)$ with lots of room.

When $N=7$, we have $\dim S = 2^3 > 7$, so $\Spin(7)$ does not embed in $SU(7)$. It does embed in $SU(8)$, though.

When $N=8$, we have $\dim S_+ = \dim S_- = 8$. Unfortunately, the corresponding maps $\Spin(8) \to SU(8)$ have kernel. Indeed, there are three maps $\Spin(8) \to SO(8)$, related by the triality symmetry of $\Spin(8)$, and each of them has a central two element group as kernel. So there are interesting maps $\Spin(8) \to SU(8)$, but no embeddings. The composition $\Spin(7) \to \Spin(8) \to SO(8)$ for one of the nononbvious maps $\Spin(8) \to SO(8)$ is the embedding of $\Spin(7)$ I referenced above.

Once $N \geq 9$, the spin representations have dimension much larger than $N$.

$\def\Spin{\mathrm{Spin}}$The spin group has a representation $S$ called the spin representation, which has dimension $2^m$ for $N=2m$ or $2m+1$. It is irreducible for $N$ odd and splits into two complex conjugate representations $S_+ \oplus S_-$ for $N$ even, with each of $S_{\pm}$ being of dimension $2^{m-1}$. These are the smallest representations of $\Spin(N)$ which do not factor through $SO(N)$. So, once the power of $2$ exceeds $N$, the spin group will not embed in $SU(N)$.

This leaves us to analyze $1 \leq N \leq 6$, and $N=8$. Each of these has some special form.

$\Spin(1)$ is the trivial group. Clearly, this embeds in $SU(1)$.

$\Spin(2)$ is the circle group $\mathrm{R}/\mathrm{Z}$. Note that $SO(2)$ and $\mathrm{Spin}(2)$ are isomorphic as abstract groups, but the map $\mathrm{Spin}(2) \to SO(2)$ is the double cover. Clearly, this embeds in $SU(2)$.

When $N=3$, the spin representation $S$ is two dimensional and gives an isomorphism $\Spin(3) \cong SU(2)$. So, in particular, $\Spin(3)$ embeds in $SU(3)$.

When $N=4$, the representations $S_+$ and $S_-$ are two dimensional, giving an isomorphism $\Spin(4) \cong SU(2) \times SU(2)$. Conveniently, this embeds into $SU(4)$ via the representation $\mathbb{C}^2 \boxtimes 1 \oplus 1 \boxtimes \mathbb{C}^2$. We cannot use just one of the two $SU(2)$ representations, as the individual projections have kernels. Particle physicists call working with $\Spin(4)$ in the form $SU(2) \times SU(2)$ the "spinor helicity formalism". If you talk to the sort of particle physicists who care what the signature of spacetime is, you'll want to know that $\Spin(3,1) \cong SL_2(\mathbb{C})$.

When $N=5$, $\dim S = 4$. Indeed, this $4$-dimensional vector space can be identified with a $2$-dimensional quaternionic vector space and $\Spin(5)$ is the group of $2 \times 2$ unitary quaternion matrices. So $\Spin(5)$ embeds into $SU(4)$ (and also into $SU(5)$.)

When $N=6$, we have $\dim S_+ = 4$ and gives an isomorphism $\Spin(6) \cong SU(4)$. So $\Spin(6)$ embeds into $SU(6)$ with lots of room.

When $N=7$, we have $\dim S = 2^3 > 7$, so $\Spin(7)$ does not embed in $SU(7)$. It does embed in $SU(8)$, though.

When $N=8$, we have $\dim S_+ = \dim S_- = 8$. Unfortunately, the corresponding maps $\Spin(8) \to SU(8)$ have kernel. Indeed, there are three maps $\Spin(8) \to SO(8)$, related by the triality symmetry of $\Spin(8)$, and each of them has a central two element group as kernel. So there are interesting maps $\Spin(8) \to SU(8)$, but no embeddings. The composition $\Spin(7) \to \Spin(8) \to SO(8)$ for one of the nononbvious maps $\Spin(8) \to SO(8)$ is the embedding of $\Spin(7)$ I referenced above.

Once $N \geq 9$, the spin representations have dimension much larger than $N$.