Here is another solution, with weaker assumptions. Suppose G tessellates a $k$-gon, where all internal vertices have valence at least 6, and the vertices of the $k$-gon have valence at least 4. Take 2 copies of G and identify the outermost $k$-cycles. This produces a planar graph where every vertex has valence at least 6, which is impossible.

Here is another solution, with weaker assumptions. Suppose G tessellates a $k$-gon, where all internal vertices have valence at least 6, and the vertices of the $k$-gon have valence at least 4. Take 2 copies of G and label the vertices of the outermost $k$-cycles as $u_1,\ldots, u_k$ and $v_1,\ldots, v_k$. Then add edges from $u_i$ to $v_i$ and from $u_i$ to $v_{i+1}$, for each $1\leq i\leq k$ and where $v_{k+1}\equiv v_1$. This produces a simple planar graph where every vertex has valence at least 6, which is impossible.