Question: In Quantum Mechanics we mostly use linear operators which often represent a physical observable,
In relativity we mostly work with tensors to describe how things change.

But both objects are (multi)-linear and map to a vector-space.

This means both objects have the same meaning but different words in different subjects but are actually equal.

Answer: 
If $V$ is a finite dimensional vector space, $\hom(V)$ is the space of linear operators on $V$ (i.e. linear maps from $V$ to itself), then $$ \hom(V)\cong V\otimes V^\ast, $$ where $V^\ast$ is the dual space of $V$ and $\cong$ denotes canonical (i.e. basis-independent) isomorphism.

If $T^{r,s}(V)=V^{\otimes r}\otimes V^{\ast\otimes s}$ denotes the space obtained by taking $r$-fold tensor products of $V$ and $s$-fold tensor product of $V^\ast$ (so-called type $(r,s)$ tensors), and define the *mixed tensor algebra* of $V$ as $$ T^{\bullet,\bullet}(V)=\bigoplus\_{r,s\in\mathbb Z^2}T^{r,s}(V), $$ this is a $\mathbb Z^2$-graded unital associative algebra whose elements (especially pure grade elements) are often referred to as *tensors over $V$*.

Thus we can say that the linear operators over $V$ are a special class of tensors, namely the $(1,1)$ grade subspace of $T^{\bullet,\bullet}(V)$.

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OP however asked about quantum mechanics, which usually involves infinite dimensional vector spaces. As always, infinite dimensional spaces have subtleties about them.

It is also possible to define tensor products of infinite dimensional vector spaces, but there are various different classes of infinite dimensional vector spaces, and the tensor products within these different categories are not the same.

To give an example relevant to quantum mechanics, if $\mathcal H$ and $\mathcal K$ are both (separable) Hilbert spaces with orthonormal Schauder bases $(\left| h\_n\right\rangle)\_{n=1,...,\infty}$ and $(\left| k\_m\right\rangle)\_{m=1,...,\infty}$, then (omitting the rigorous definition), the Hilbert space tensor product $\mathcal H\otimes\mathcal K$ is defined in such a way that the system $$ (\left| h\_n\right\rangle\otimes\left| k\_m\right\rangle)\_{m,n=1,...,\infty} $$ is an orthonormal Schauder basis of the Hilbert space $\mathcal H\otimes\mathcal K$.

For simplicity consider real Hilbert spaces because I don't want to deal with complex conjugations and antilinear things. A Hilbert space is always isomorphic to its topological dual ($\mathcal H\cong\mathcal H^\sharp$), and I will make this identification.

One may thus try to identify $$ \mathcal {H}\otimes\mathcal {H}^\sharp\cong\mathcal H\otimes\mathcal H$$ with the space $\hom(\mathcal H)$ of linear operators on $\mathcal H$.

However since $\mathcal H\otimes\mathcal H$ is a Hilbert space, if we write $$ A=\sum\_{n,m}a\_{nm}\left| h\_n\right\rangle \otimes \left| k\_m\right\rangle,\quad B=\sum\_{n,m}b\_{nm}\left| h\_n\right\rangle \otimes \left| k\_m\right\rangle, $$ the induced inner product is finite and is $$ \langle A,B\rangle=\sum\_{n,m}a\_{nm}b\_{nm}<\infty. $$

Thus elements of $\mathcal H\otimes\mathcal H$ are [Hilbert-Schmidt operators](https://en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator), but not all relevant operators in QM are Hilbert-Schmidt.

Thus we see that $\hom(\mathcal H)$ is not equivalent with the *Hilbert space* tensor product $\mathcal H\otimes\mathcal H$.
A key difference between the linear operators of QM and the tensors of GR is the structure of the underlying vector spaces. The vector spaces on which QM operators act are complex Hilbert spaces, typically with infinite dimensions (although simple examples such as a particle with spin $\pm 1$ may be finite dimensional). GR tensors on the other hand act on vectors and vector fields in spacetime which has just four real dimensions.